CP2 §7.3 Second-order non-homogeneous differential equations

Core pure mathematics 2

Table of contents

1. Complementary function + Particular integral
2. An example
3. Another example
4. Yet another example
5. An investigation: more general cases
6. Exercise 7C

1. Complementary function + Particular integral

Second-order differentail equations of the form $$ \begin{align} a \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } + b \frac{ \textrm{d}y }{ \textrm{d}x } + c y = \textrm f(x) \end{align} $$ are non-homogeneous.

 

(i) To solve an equation of this type, we first find the general solution of the corresponding homogeneous differential equation, $$ \begin{align} a \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } + b \frac{ \textrm{d}y }{ \textrm{d}x } + c y = 0. \end{align} $$ This is called the complementary function (C.F.).

 

(ii) We then need to find a particular integral (P.I.), which is a function that satisfies the original differential equation with $\textrm f(x)$ on the right-hand side. The form of the particular integral depends on the form of $ \textrm f(x) $.

 

This table provides some particular integrals to try. We substitute the P.I. into the original equation to find the values of the coefficients such as $\lambda, \mu$ and $\nu$.

 

$\textrm f(x)$	Form of the particular integral
$p$	$\lambda$
$p + qx$	$\lambda + \mu x$
$p + qx + rx^2$	$\lambda + \mu x + \nu x^2$
$p \textrm{e}^{kx}$	$\lambda \textrm{e}^{kx}$
$p \cos \omega x + q \sin \omega x$	$\lambda \cos \omega x + \mu \sin \omega x$

 

(iii) The general solution to the differential equation is: $$ \begin{align} y = \textrm{C.F.} + \textrm{P.I.} \end{align} $$ since (let $ y_1 = \textrm{C.F.} $ and $ y_2 = \textrm{P.I.} $) $$ \begin{align} \left( a \frac{ \textrm{d}^2 }{ \textrm{d}x^2 } + b \frac{ \textrm{d} }{ \textrm{d}x } + c \right) y &= \left( a \frac{ \textrm{d}^2 }{ \textrm{d}x^2 } + b \frac{ \textrm{d} }{ \textrm{d}x } + c \right) ( y_1 + y_2 ) \\ &= \underbrace{ \left( a \frac{ \textrm{d}^2 }{ \textrm{d}x^2 } + b \frac{ \textrm{d} }{ \textrm{d}x } + c \right) y_1 }_{ = 0} + \underbrace{ \left( a \frac{ \textrm{d}^2 }{ \textrm{d}x^2 } + b \frac{ \textrm{d} }{ \textrm{d}x } + c \right) y_2 }_{ = \textrm f(x) } = \textrm f(x). \qquad \square \end{align} $$

2. An example

Example 1. Consider $$ \begin{align} \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } - 5 \frac{ \textrm{d}y }{ \textrm{d}x } + 6 y = \textrm f(x) \end{align} $$ (i) Find the complementary function.

 

(ii) Find a particular integral of the differential equation, and thus the general solution, when $ \textrm f(x) $ is: $$ \begin{align} \textbf{(a)} \quad 3 \qquad \textbf{(b)} \quad 2x \qquad \textbf{(c)} \quad 3x^2 \qquad \textbf{(d)} \quad \textrm e^x \qquad \textbf{(e)} \quad 13 \sin 3x \end{align} $$

 

Solution. (i) For the complementary function, we have: $$ \begin{align} \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } - 5 \frac{ \textrm{d}y }{ \textrm{d}x } + 6 y = 0 \end{align} $$ For $y = A \textrm e^{ mx }$, the auxiliary equation reads $$ \begin{align} && m^2 - 5m + 6 &= 0 \\ &\Rightarrow & ( m - 2 ) ( m - 3 ) & = 0 \\ &\Rightarrow & m = 2 \quad \textrm{or} \quad m &= 3 \end{align} $$ and the complementary function is $$ \begin{align} y_1 = A \textrm e^{ 2x } + B \textrm e^{ 3x } \end{align} $$

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(ii) (a) $\textrm f(x)=3$.

 

Let $ y = \lambda $. Substituting this into the differential equation gives

$$ \begin{align} 6 \lambda = 3 \qquad \Rightarrow \qquad \lambda = \frac12 \end{align} $$ The general solution reads $$ \begin{align} \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } - 5 \frac{ \textrm{d}y }{ \textrm{d}x } + 6 y = 3 \qquad \Rightarrow \qquad y = A \textrm e^{ 2x } + B \textrm e^{ 3x } + \frac12 \end{align} $$

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(b) $\textrm f(x)=2x$.

 

Let $ y = \lambda + \mu x $. $$ \begin{align} y &= \lambda + \mu x \\ y' &= \mu \\ y'' &= 0 \end{align} $$ Substituting these into the differential equation gives $$ \begin{align} y'' - 5 y' + 6y &= -5 \mu + 6 ( \lambda + \mu x ) \\ &= ( 6 \lambda - 5 \mu ) + 6 \mu x \stackrel{!}{=} 2x \end{align} $$ i.e. $$ \begin{align} \left\{ \begin{array}{ll} 6 \mu = 2 \\ 6 \lambda - 5 \mu = 0 \end{array} \right. \qquad \Rightarrow \qquad \left\{ \begin{array}{ll} \mu = \frac13 \\ \lambda = \frac{ 5 \mu }{ 6 } = \frac{ 5 \times \frac13 }{ 6 } = \frac{ 5 }{ 18 } \end{array} \right. \end{align} $$ So the particular integral is $$ \begin{align} y = \frac{ 5 }{ 18 } + \frac13x \end{align} $$ and the general solution is $$ \begin{align} \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } - 5 \frac{ \textrm{d}y }{ \textrm{d}x } + 6 y = 2x \qquad \Rightarrow \qquad y = \textrm{C.F.} + \textrm{P.I.} = A \textrm e^{ 2x } + B \textrm e^{ 3x } + \frac{ 5 }{ 18 } + \frac13x \end{align} $$

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(c) $\textrm f(x)= 3x^2$.

 

Let $ y = \lambda + \mu x + \nu x^2 $.

$$ \begin{align} y &= \lambda + \mu x + \nu x^2 \\ y' &= \mu + 2 \nu x \\ y'' &= 2 \nu \end{align} $$ Substituting these into the differential equation gives $$ \begin{align} y'' - 5 y' + 6y &= 2 \nu - 5 ( \mu + 2 \nu x) + 6 \left( \lambda + \mu x + \nu x^2 \right) \\ &= ( 6 \lambda - 5 \mu + 2 \nu ) + ( 6 \mu - 10 \nu ) x + 6 \nu x^2 \stackrel{!}{=} 3x^2 \end{align} $$ i.e. $$ \begin{align} \left\{ \begin{array}{ll} 6 \nu = 3 \\ 6 \mu - 10 \nu = 0 \\ 6 \lambda - 5 \mu + 2 \nu = 0 \end{array} \right. \qquad \Rightarrow \qquad \left\{ \begin{array}{ll} \nu = \frac12 \\ \mu = \frac{ 5 \nu }{ 3 } = \frac{ 5 \times \frac12 }{ 3 } = \frac{ 5 }{ 6 } \\ \lambda = \frac{ 5 \mu - 2 \nu }{ 6 } = \frac{ 5 \times \frac56 - 2 \times \frac12 }{ 6 } = \frac{ \frac{19}{6} }{6} = \frac{19}{36} \end{array} \right. \end{align} $$ So the particular integral is $$ \begin{align} y = \frac{19}{36} + \frac56 x + \frac12 x^2 \end{align} $$ and the general solution is $$ \begin{align} \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } - 5 \frac{ \textrm{d}y }{ \textrm{d}x } + 6 y = 3x^2 \qquad \Rightarrow \qquad y = \textrm{C.F.} + \textrm{P.I.} = A \textrm e^{ 2x } + B \textrm e^{ 3x } + \frac{19}{36} + \frac56 x + \frac12 x^2 \end{align} $$

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(d) $\textrm f(x) = \textrm e^x$.

 

Let $ y = \lambda \textrm e^{ x } $.

$$ \begin{align} y &= \lambda \textrm e^{ x } \\ y' &= \lambda \textrm e^{ x } \\ y'' &= \lambda \textrm e^{ x } \end{align} $$ Substituting these into the differential equation gives $$ \begin{align} y'' - 5 y' + 6y &= ( 1 - 5 + 6 ) \lambda \textrm e^{ x } = 2 \lambda \textrm e^{ x } \stackrel{!}{=} \textrm e^x \qquad \Rightarrow \qquad \lambda = \frac12 \end{align} $$ So the particular integral is $$ \begin{align} y = \frac12 \textrm e^{ x } \end{align} $$ and the general solution is $$ \begin{align} \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } - 5 \frac{ \textrm{d}y }{ \textrm{d}x } + 6 y = \textrm e^x \qquad \Rightarrow \qquad y = \textrm{C.F.} + \textrm{P.I.} = A \textrm e^{ 2x } + B \textrm e^{ 3x } + \frac12 \textrm e^x \end{align} $$

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(e) $\textrm f(x)=13 \sin 3x$.

 

Let $ y = \lambda \cos 3x + \mu \sin 3x $.

$$ \begin{align} y &= \lambda \cos 3x + \mu \sin 3x \\ y' &= - 3 \lambda \sin 3x + 3 \mu \cos 3x \\ y'' &= - 9 \lambda \cos 3x - 9 \mu \sin 3x \end{align} $$ Substituting these into the differential equation gives $$ \begin{align} y'' - 5 y' + 6y &= ( - 9 \lambda \cos 3x - 9 \mu \sin 3x ) \\ &\quad - 5 ( - 3 \lambda \sin 3x + 3 \mu \cos 3x ) \\ &\quad + 6 ( \lambda \cos 3x + \mu \sin 3x ) \\ &= ( -3 \lambda - 15 \mu ) \cos 3x + ( -3 \mu + 15 \lambda ) \sin 3x \stackrel{!}{=} 13 \sin 3x \end{align} $$ i.e. $$ \begin{align} \left\{ \begin{array}{ll} \lambda + 5 \mu = 0 \\ 15 \lambda - 3 \mu = 13 \end{array} \right. \qquad \Rightarrow \qquad \left\{ \begin{array}{ll} \mu = - \frac{13}{78} = -\frac16 \\ \lambda = - 5 \mu = \frac56 \end{array} \right. \end{align} $$ So the particular integral is $$ \begin{align} y = \frac56 \cos 3x - \frac16 \sin 3x \end{align} $$ and the general solution is $$ \begin{align} \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } - 5 \frac{ \textrm{d}y }{ \textrm{d}x } + 6 y = 13 \sin 3x \qquad \Rightarrow \qquad y = \textrm{C.F.} + \textrm{P.I.} = A \textrm e^{ 2x } + B \textrm e^{ 3x } + \frac56 \cos 3x - \frac16 \sin 3x \end{align} $$

3. Another example

We need to be careful if the standard form of the particular contains terms which form part of the complementary function. If this is the case, we need to modify the particular integral so that no two terms in the general solution have the same form.

 

For example, this situation occurs when $ \textrm f(x) $ is of the form $ p \textrm e^{ k x } $, and $k$ is one of the roots of the auxiliary equation. In this case, we can try a particular integral of the form, $ \lambda x \textrm e^{ k x }$.

 

Example 2. Find the general solution to the differential equation $$ \begin{align} \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } - 5 \frac{ \textrm{d}y }{ \textrm{d}x } + 6 y = \textrm e^{ 2 x } \end{align} $$

 

Solution.

 

(i) Complementary function: Let $ y = A \textrm e^{ mx }$ and the auxiliary equation reads $$ \begin{align} && m^2 - 5m + 6 &= 0 &\Rightarrow& ( m - 2 )( m - 3 ) &= 0 \\ &\Rightarrow& m = 2 \quad \textrm{or} \quad m&=3 \end{align} $$ and the complementary function is $$ \begin{align} y_1 = A \textrm e^{ 2x } + B \textrm e^{ 3x } \end{align} $$

 

(ii) Particular integral: The standard form, $\lambda \textrm e^{ 2x }$, is already part of the C.F. so we try the standard form multiplied by $x$, i.e. $$ \begin{align} y_2 &= \lambda x \textrm e^{ 2x } \\ y_2' &= \lambda \textrm e^{ 2x } + 2 \lambda x \textrm e^{ 2x } \\ &= (1 + 2x) \lambda \textrm e^{ 2x } \\ y_2'' &= 2 \lambda \textrm e^{ 2x } + 2 (1+2x) \lambda \textrm e^{ 2x } \\ &= (4 + 4x) \lambda \textrm e^{ 2x } \end{align} $$ Substituting these into the differential equation gives $$ \begin{align} y_2'' - 5y_2' + 6y_2 &= (4 + 4x) \lambda \textrm e^{ 2x } - 5 (1 + 2x) \lambda \textrm e^{ 2x } + 6 \lambda x \textrm e^{ 2x } \\ &= - \lambda \textrm e^{ 2x } + \underbrace{ (4 - 10 + 6 ) }_{=0} \lambda x \textrm e^{ 2x } \\ &\stackrel{!}{=} \textrm e^{ 2x } \end{align} $$ We find: $$ \begin{align} \lambda = -1 \end{align} $$ and the particular integral is $$ \begin{align} y_2 &= - x \textrm e^{ 2x } \end{align} $$ Finally, the general solution reads $$ \begin{align} y = y_1 + y_2 = A \textrm e^{ 2x } + B \textrm e^{ 3x } - x \textrm e^{ 2x } \qquad \checkmark \end{align} $$

4. Yet another example

When one of the roots of the auxiliary equation is 0, the complementary function will contain a constant term. If $ \textrm f(x) $ is a polynomial, we will need to multiply its particular integral by $x$ to make sure the P.I. does not also contain a constant term.

 

Example 3. Find the general solution to the differential equation $$ \begin{align} \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } - 2 \frac{ \textrm{d}y }{ \textrm{d}x } = 3 \end{align} $$

 

Solution.

 

(i) Complementary function: Let $ y = A \textrm e^{ mx }$ and the auxiliary equation reads $$ \begin{align} && m^2 - 2m &= 0 &\Rightarrow& m( m - 2 ) &= 0 \\ &\Rightarrow& m = 0 \quad \textrm{or} \quad m&=2 \end{align} $$ and, noting that $A \textrm e^{0x} = A$, the complementary function is $$ \begin{align} y_1 = A + B \textrm e^{ 2x } \end{align} $$

 

(ii) Particular integral: The standard form, $y_2 = \lambda$, is already part of the C.F. so we try the standard form multiplied by $x$, i.e. $$ \begin{align} y_2 &= \lambda x \\ y_2' &= \lambda \\ y_2'' &= 0 \end{align} $$ Substituting these into the differential equation gives $$ \begin{align} y_2'' - 2y_2' &= -2 \lambda \stackrel{!}{=} 3 \end{align} $$ We find: $$ \begin{align} \lambda = - \frac32 \end{align} $$ and the particular integral is $$ \begin{align} y_2 &= - \frac32 x \end{align} $$ Finally, the general solution reads $$ \begin{align} y = y_1 + y_2 = A + B \textrm e^{ 2x } - \frac32 x \qquad \checkmark \end{align} $$

5. An investigation: More general cases

Question 1. More generally, we consider $$ \begin{align} a \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } + b \frac{ \textrm{d}y }{ \textrm{d}x } + c y = \textrm f(x) \end{align} $$

where

$$ \begin{align} \textbf{(a)} &&& \textrm f(x) = p \\ \\ \textbf{(b)} &&& \textrm f(x) = p + q x \\ \\ \textbf{(c)} &&& \textrm f(x) = p + q x + r x^2 \\ \\ \textbf{(d)} &&& \textrm f(x) = p \textrm e^{ k x } \\ \\ \textbf{(e)} &&& \textrm f(x) = p \cos \omega x + q \sin \omega x \end{align} $$

 

For the complementary function, the auxiliary equation reads $$ \begin{align} am^2 + bm + c \qquad \Rightarrow \qquad m = \frac{ - b \pm \sqrt{ b^2 - 4ac } }{ 2a } \end{align} $$

 

(a) $ \textrm f(x) = p $.

Case 1. $c \ne 0$:Let $ y = \lambda $. Substituting this into the differential equation gives

$$ \begin{align} a y'' + b y' + c y = c \lambda \stackrel{!}{=} p \qquad \Rightarrow \qquad \lambda = \frac{p}{c} \qquad (c \ne 0) \end{align} $$

Case 2. $c=0$: The auxiliary equation reads $$ \begin{align} am^2 + bm = 0 \qquad \Rightarrow \qquad m = 0, \quad -\frac{b}{a} \end{align} $$ so the complementary function is $$ \begin{align} y_1 = A + B \textrm e^{ - \frac{b}{a} x } \end{align} $$ Since the C.F. already constains a constant term, we try a particular integral of the form $$ \begin{align} y &= \lambda x \\ y' &= \lambda \\ y'' &= 0 \end{align} $$ and find: $$ \begin{align} a y'' + b y' &= b \lambda \stackrel{!}{=} p \qquad \Rightarrow \qquad \lambda = \frac{p}{b} \end{align} $$ So the general solution reads $$ \begin{align} & a \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } + b \frac{ \textrm{d}y }{ \textrm{d}x } + c y = p & &\Rightarrow& y &= A \textrm e^{ m_1 x } + B \textrm e^{ m_2 x } + \frac{p}{c} \\ \\ & a \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } + b \frac{ \textrm{d}y }{ \textrm{d}x } = p & &\Rightarrow& y &= A + B \textrm e^{ - \frac{b}{a} x } + \frac{p}{b} x \end{align} $$

 

(b) $ \textrm f(x) = p + qx $.

Case 1. $c \ne 0$: Let $ y = \lambda + \mu x $.

$$ \begin{align} y &= \lambda + \mu x \\ y' &= \mu \\ y'' &= 0 \end{align} $$ Substituting these into the differential equation gives $$ \begin{align} a y'' + b y' + c y &= b \mu + c ( \lambda + \mu x ) \\ &= ( b \mu + c \lambda ) + c \mu x \stackrel{!}{=} p + qx \end{align} $$ i.e. $$ \begin{align} \left\{ \begin{array}{ll} c \mu = q \\ b \mu + c \lambda = p \end{array} \right. \qquad \Rightarrow \qquad \left\{ \begin{array}{ll} \mu = \frac{q}{c} \\ \lambda = \frac{ p - b \mu }{ c } = \frac{ p - b \times \frac{q}{c} }{ c } = \frac{ cp - bq }{ c^2 } \end{array} \right. \end{align} $$ So the particular integral is $$ \begin{align} y = \frac{ cp - bq }{ c^2 } + \frac{q}{c} x \qquad ( c \ne 0 ) \end{align} $$

Case 2. $c = 0$: The auxiliary equation reads $$ \begin{align} am^2 + bm = 0 \qquad \Rightarrow \qquad m = 0, \quad -\frac{b}{a} \end{align} $$ so the complementary function is $$ \begin{align} y_1 = A + B \textrm e^{ - \frac{b}{a} x } \end{align} $$ Since the C.F. already constains a constant term, we try a particular integral of the form $$ \begin{align} y &= \lambda x + \mu x^2 \\ y' &= \lambda + 2 \mu x \\ y'' &= 2 \mu \end{align} $$ and find: $$ \begin{align} a y'' + b y' &= 2a \mu + b ( \lambda + 2 \mu x ) \\ &= (2a \mu + b \lambda) + 2 b \mu \stackrel{!}{=} p + qx \end{align} $$ i.e. $$ \begin{align} & \Rightarrow & &\left\{ \begin{array}{ll} 2 b \mu = q \\ 2a \mu + b \lambda = p \end{array} \right. \\ \\ & \Rightarrow & &\left\{ \begin{array}{ll} \mu = \frac{q}{2b} \\ \lambda = \frac{ p - 2a \mu }{ b } = \frac{ p - 2a \left( \frac{q}{2b} \right) }{ b } = \frac{ b p - a q }{ b^2 } \end{array} \right. \end{align} $$ and the particular integral is $$ \begin{align} y = \frac{ b p - a q }{ b^2 } x + \frac{q}{2b} x^2 \end{align} $$ So the general solution reads $$ \begin{align} & a \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } + b \frac{ \textrm{d}y }{ \textrm{d}x } + c y = p + qx & &\Rightarrow& y &= A \textrm e^{ m_1 x } + B \textrm e^{ m_2 x } + \frac{ cp - bq }{ c^2 } + \frac{q}{c} x \\ \\ & a \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } + b \frac{ \textrm{d}y }{ \textrm{d}x } = p + qx & &\Rightarrow& y &= A + B \textrm e^{ - \frac{b}{a} x } + \frac{ b p - a q }{ b^2 } x + \frac{q}{2b} x^2 \end{align} $$

 

(c) $ \textrm f(x) = p + qx + rx^2 $.

Case 1. $c \ne 0$: Let $ y = \lambda + \mu x + \nu x^2 $.

$$ \begin{align} y &= \lambda + \mu x + \nu x^2 \\ y' &= \mu + 2 \nu x \\ y'' &= 2 \nu \end{align} $$ Substituting these into the differential equation gives $$ \begin{align} a y'' + b y' + c y &= 2a \nu + b ( \mu + 2 \nu x) + c \left( \lambda + \mu x + \nu x^2 \right) \\ &= ( c \lambda + 2 b \mu + 2 a \nu ) + ( c \mu + 2 b \nu ) x + c \nu x^2 \stackrel{!}{=} p + qx + rx^2 \end{align} $$ i.e. $$ \begin{align} & \Rightarrow & &\left\{ \begin{array}{ll} c \nu = p \\ c \mu + 2 b \nu = q \\ c \lambda + 2 b \mu + 2 a \nu = r \end{array} \right. \\ \\ & \Rightarrow & &\left\{ \begin{array}{ll} \nu = \frac{p}{c} \\ \mu = \frac{ q - 2 b \nu }{ c } = \frac{ q - 2 b \left( \frac{p}{c} \right) }{ c } = \frac{ c q - 2 b p }{ c^2 } \\ \lambda = \frac{ r - 2 b \mu - 2 a \nu }{ c } = \frac{ r - 2 b \mu - 2 a \nu }{ c } = \frac{ r - 2b \left( \frac{ c q - 2 b p }{ c^2 } \right) - 2a \left( \frac{p}{c} \right) }{ c } = \frac{ r c^2 - 2b ( c q - 2 b p ) - 2ac p }{ c^3 } = \frac{ 2 \left( 2 b^2 - a c \right) p - 2 b c q + c^2 r } { c^3 } \end{array} \right. \end{align} $$ So the particular integral is $$ \begin{align} y = \frac{ 2 \left( 2 b^2 - a c \right) p - 2 b c q + c^2 r } { c^3 } + \left( \frac{ c q - 2 b p }{ c^2 } \right) x + \frac{p}{c} x^2 \qquad ( c \ne 0 ) \end{align} $$

Case 2. $c = 0$: The auxiliary equation reads $$ \begin{align} am^2 + bm = 0 \qquad \Rightarrow \qquad m = 0, \quad -\frac{b}{a} \end{align} $$ so the complementary function is $$ \begin{align} y_1 = A + B \textrm e^{ - \frac{b}{a} x } \end{align} $$ Since the C.F. already constains a constant term, we try a particular integral of the form $$ \begin{align} y &= \lambda x + \mu x^2 + \nu x^3 \\ y' &= \lambda + 2 \mu x + 3 \nu x^2 \\ y'' &= 2 \mu + 6 \nu x \end{align} $$ and find: $$ \begin{align} a y'' + b y' &= a ( 2 \mu + 6 \nu x ) + b \left( \lambda + 2 \mu x + 3 \nu x^2 \right) \\ &= (2 a \mu + b \lambda ) + ( 6 a \nu + 2 b \mu ) x + 3 b \nu x^2 \stackrel{!}{=} p + qx + rx^2 \end{align} $$ i.e. $$ \begin{align} & \Rightarrow & &\left\{ \begin{array}{ll} 3 b \nu = r \\ 6 a \nu + 2 b \mu = q \\ 2 a \mu + b \lambda = p \end{array} \right. \\ \\ & \Rightarrow & &\left\{ \begin{array}{ll} \nu = \frac{ r }{ 3b } \\ \mu = \frac{ q - 6a \nu }{ 2b } = \frac{ q - 6a \left( \frac{ r }{ 3b } \right) }{ 2b } = \frac{ b q - 2 a r }{ 2 b^2 } \\ \lambda = \frac{ p - 2a \mu }{ b } = \frac{ p - 2a \left( \frac{ b q - 2 a r }{ 2 b^2 } \right) }{ b } = \frac{ b^2 p - a ( b q - 2 a r ) }{ b^3 } = \frac{ b^2 p - a b q + 2 a^2 r ) }{ b^3 } \end{array} \right. \end{align} $$ and the particular integral is $$ \begin{align} y = \frac{ b^2 p - a b q + 2 a^2 r ) }{ b^3 } x + \frac{ b q - 2 a r }{ 2 b^2 } x^2 + \frac{ r }{ 3b } x^3 \end{align} $$ So the general solution reads $$ \begin{align} & a \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } + b \frac{ \textrm{d}y }{ \textrm{d}x } + c y = p + qx & &\Rightarrow& y &= A \textrm e^{ m_1 x } + B \textrm e^{ m_2 x } + \frac{ 2 \left( 2 b^2 - a c \right) p - 2 b c q + c^2 r } { c^3 } + \left( \frac{ c q - 2 b p }{ c^2 } \right) x + \frac{p}{c} x^2 \\ \\ & a \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } + b \frac{ \textrm{d}y }{ \textrm{d}x } = p + qx & &\Rightarrow& y &= A + B \textrm e^{ - \frac{b}{a} x } + \frac{ b^2 p - a b q + 2 a^2 r ) }{ b^3 } x + \frac{ b q - 2 a r }{ 2 b^2 } x^2 + \frac{ r }{ 3b } x^3 \end{align} $$

 

Summary. We notice that, if $c=0$, the C.F. contains a constant term. If, in addition, $\textrm f(x)$ is a polynomial, since the constant term is already part of the C.F., we should multiply the standard form of the particular integral by $x$, as illustrated above and summarised below.

 

$\textrm f(x)$	Form of the particular integral
$p$	$\lambda x$
$p + qx$	$\lambda x + \mu x^2$
$p + qx + rx^2$	$\lambda x + \mu x^2 + \nu x^3$

 

Aside. Proof using the integrating factor:

We can alternatively prove the result summarised in the table by regarding the differential equation as essentially a 'first-order DE' for $y'$, i.e. $$ \begin{align} && a \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } + b \frac{ \textrm{d}y }{ \textrm{d}x } = a \frac{ \textrm{d}y' }{ \textrm{d}x } + b y' &= \textrm f(x) \\ \\ &\Rightarrow& \frac{ \textrm{d}y' }{ \textrm{d}x } + \frac{b}{a} y' &= \frac1a \textrm f(x) \end{align} $$ The integrating factor reads $$ \begin{align} \textrm e^{ \int \frac{b}{a} \, \textrm dx } = \textrm e^{ \frac{b}{a} x } \end{align} $$ As we multiply the differential equation by the integrating factor, $$ \begin{align} \textrm e^{ \frac{b}{a} x } \frac{ \textrm{d}y' }{ \textrm{d}x } + \frac{b}{a} \textrm e^{ \frac{b}{a} x } y' = \frac{ \textrm{d} }{ \textrm{d}x } \left[ \textrm e^{ \frac{b}{a} x } y' \right] &= \frac1a \textrm f(x) \textrm e^{ \frac{b}{a} x } \\ \\ \Rightarrow \qquad \textrm e^{ \frac{b}{a} x } y' = \frac1a \int \left[ \textrm f(x) \textrm e^{ \frac{b}{a} x } \right] \, \textrm dx \end{align} $$ Suppose that $\textrm f(x)$ is a polynomial of the form, $$ \begin{align} \textrm f(x) = \sum_{n=0}^N a_n x^n \end{align} $$ Then we have: $$ \begin{align} \Rightarrow \qquad \textrm e^{ \frac{b}{a} x } y' = \frac1a \sum_{n=0}^N a_n \int \left[ x^n \textrm e^{ \frac{b}{a} x } \right] \, \textrm dx \end{align} $$ Recall from FP2 §6.1 Reduction formulae, $$ \begin{align} \int x^n \textrm e^{kx} \, \textrm dx &= \frac1k x^n \textrm e^{kx} - \frac{ n }{ k^2 } x^{n-1} \textrm e^{kx} + \frac{ n(n-1) }{ k^3 } x^{n-2} \textrm e^{kx} - \frac{ n(n-1)(n-2) }{ k^4 } x^{n-3} \textrm e^{kx} \\ &\qquad + \cdots + (-1)^n \frac{ n(n-1)(n-2)\cdots1 }{ k^{n+1} } \textrm e^{kx} \\ &= \sum_{ r = 0 }^{ n } (-1)^r \frac{ n! }{ (n-r)! } \frac{ 1 }{ k^{r+1} } x^{n-r} \textrm e^{kx} + C \\ &\equiv \sum_{r = 0}^{ n } b_r x^r \textrm e^{kx} + C \end{align} $$ This yields $$ \begin{align} & \Rightarrow & \textrm e^{ \frac{b}{a} x } y' &= \frac1a \sum_{n=0}^N a_n \left( \sum_{r = 0}^{ n } b_r x^r \textrm e^{ \frac{b}{a}x } + C \right) \\ \\ & \Rightarrow & y' &= \frac{C}{a} \textrm e^{ - \frac{b}{a} x } + \frac1a \sum_{n=0}^N a_n \left( \sum_{r = 0}^{ n } b_r x^r \right) \\ \\ & \Rightarrow & y &= - \frac{C}{b} \textrm e^{ - \frac{b}{a} x } + D + \underbrace{ \frac1a \sum_{n=0}^N a_n \left( \sum_{r = 0}^{ n } \frac{b_r}{r+1} x^{r+1} \right) }_{ = c_1 x + c_2 x^2 + \cdots + c_{N+1} x^{N+1} = \textrm{P.I.}} \end{align} $$ where $C$ and $D$ are arbitrary constants. We see that, for $c=0$, if $\textrm f(x)$ is a polynomial of order $N$, the particular integral is a polynomial of order $N+1$ without a constant term, i.e. it is the standard form multiplied by $x. \qquad \square$

 

(d) $ \textrm f(x) = p \textrm e^{ k x } $.

 

Case 1. $ ak^2 + bk + c \ne 0$: i.e. $\textrm e^{ k x }$ is not part of the complementary function. Let $ y = \lambda \textrm e^{ k x } $.

$$ \begin{align} y &= \lambda \textrm e^{ k x } \\ y' &= \lambda k \textrm e^{ k x } \\ y'' &= \lambda k^2 \textrm e^{ k x } \end{align} $$ Substituting these into the differential equation gives $$ \begin{align} a y'' + b y' + c y &= ( a k^2 + b k + c ) \lambda \textrm e^{ k x } \stackrel{!}{=} p \textrm e^{ k x } \qquad \Rightarrow \qquad \lambda = \frac{ p }{ a k^2 + b k + c } \end{align} $$ So the particular integral is $$ \begin{align} y = \frac{ p }{ a k^2 + b k + c } \textrm e^{ k x } \end{align} $$

 

Case 2. $ ak^2 + bk + c = 0$ and $b^2-4ac \ne 0$: i.e. $\textrm e^{ k x }$ is part of the complementary function.

Let $ y = \lambda x \textrm e^{ k x } $. $$ \begin{align} y &= \lambda x \textrm e^{ k x } \\ y' &= \lambda \textrm e^{ k x } + \lambda k x \textrm e^{ k x } = ( 1 + k x ) \lambda \textrm e^{ k x } \\ y'' &= \lambda k \textrm e^{ k x } + \left( k + k^2 x \right) \lambda \textrm e^{ k x } = \left( 2k + k^2 x \right) \lambda \textrm e^{ k x } \end{align} $$ Substituting these into the differential equation gives $$ \begin{align} a y'' + b y' + c y &= a \left( 2k + k^2 x \right) \lambda \textrm e^{ k x } + b ( 1 + k x ) \lambda \textrm e^{ k x } + c \lambda x \textrm e^{ k x } \\ &= (2ak + b) \lambda \textrm e^{ k x } + \underbrace{ \left( a k^2 + b k + c \right) }_{ = 0 } \lambda x \textrm e^{ k x } \\ &\stackrel{!}{=} p \textrm e^{ k x } \\ \\ \Rightarrow \qquad \lambda &= \frac{ p }{ 2ak + b } \end{align} $$ so the particular integral is $$ \begin{align} y = \frac{ p }{ 2ak + b } x \textrm e^{ k x } \end{align} $$ The general solution reads $$ \begin{align} a \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } + b \frac{ \textrm{d}y }{ \textrm{d}x } + c y = p \textrm e^{ k x } \qquad \Rightarrow \qquad y = \left\{ \begin{array}{lcl} A \textrm e^{ m_1 x } + B \textrm e^{ m_2 x } + \left( \frac{ p }{ a k^2 + b k + c } \right) \textrm e^{ k x } & \textrm{for} & ak^2 + bk + c \ne 0 \\ A \textrm e^{ m x } + \left( B + \frac{ p }{ 2ak + b } x \right) \textrm e^{ k x } & \textrm{for} & ak^2 + bk + c = 0 \end{array} \right. \end{align} $$

 

Aside. Do we not need to check any other terms of the form $x^n \textrm e^{ k x }$?

(i) Step 1. We show the absence of the $ x^2 \textrm e^{ k x } $ terms. $$ \begin{align} y &= \mu x^2 \textrm e^{ k x } \\ y' &= 2 \mu x \textrm e^{ k x } + k \mu x^2 \textrm e^{ k x } \\ &= \left( 2 x + k x^2 \right) \mu \textrm e^{ k x } \\ y'' &= ( 2 + 2 k x ) \mu \textrm e^{ k x } + k \left( 2 x + k x^2 \right) \mu \textrm e^{ k x } \\ &= \left( 2 + 4 k x + k^2 x^2 \right) \mu \textrm e^{ k x } \end{align} $$ Substituting these into the differential equation gives $$ \begin{align} a y'' + b y' + c y &= a \left( 2 + 4 k x + k^2 x^2 \right) \mu \textrm e^{ k x } \\ &\qquad + b \left( 2 x + k x^2 \right) \mu \textrm e^{ k x } \\ &\qquad + c x^2 \mu \textrm e^{ k x } \\ &= \Big[ 2a + 2 ( 2 a k + b ) x + \underbrace{ \left( a k ^2 + b k + c \right) }_{ = 0 } x^2 \Big] \mu \textrm e^{ k x } \\ &= \Big[ 2a + 2 ( 2 a k + b ) x \Big] \mu \textrm e^{ k x } \\ &\stackrel{!}{=} p \textrm e^{ k x } \end{align} $$ and the absence of $x$-terms on the right-hand side requires $\mu = 0$.

 

(ii) Step 2. Let's combine $x \textrm e^{ k x }$ and $x^2 \textrm e^{ k x }$ terms together. $$ \begin{align} y &= \lambda x \textrm e^{ k x } + \mu x^2 \textrm e^{ k x } \\ y' &= ( 1 + k x ) \lambda \textrm e^{ k x } + \left( 2 x + k x^2 \right) \mu \textrm e^{ k x } \\ &= \Big[ ( \lambda + ( k \lambda + 2 \mu ) x + k \mu x^2 \Big] \textrm e^{ k x } \\ y'' &= \left( 2k + k^2 x \right) \lambda \textrm e^{ k x } + \left( 2 + 4 k x + k^2 x^2 \right) \mu \textrm e^{ k x } \\ &= \Big[ ( 2 k \lambda + 2 \mu ) + \left( k^2 \lambda + 4 k \mu \right) x + k^2 \mu x^2 \Big] \textrm e^{ k x } \end{align} $$ Substituting these into the differential equation gives $$ \begin{align} a y'' + b y' + c y &= a \Big[ ( 2 k \lambda + 2 \mu ) + \left( k^2 \lambda + 4 k \mu \right) x + k^2 \mu x^2 \Big] \textrm e^{ k x } \\ &\qquad + b \Big[ ( \lambda + ( k \lambda + 2 \mu ) x + k \mu x^2 \Big] \textrm e^{ k x } \\ &\qquad + c \Big[ \lambda x + \mu x^2 \Big] \textrm e^{ k x } \\ &= \Big[ (2ak + b) \lambda + 2a \mu + \underbrace{ \left( a k^2 + b k + c \right) }_{ = 0 } \lambda x + 2 ( 2 a k + b ) \mu x + \underbrace{ \left( a k ^2 + b k + c \right) }_{ = 0 } \mu x^2 \Big] \textrm e^{ k x } \\ &= \Big[ (2ak + b) \lambda + 2a \mu + 2 ( 2 a k + b ) \mu x \Big] \textrm e^{ k x } \\ &\stackrel{!}{=} p \textrm e^{ k x } \end{align} $$ i.e. $$ \begin{align} \left\{ \begin{array}{l} ( 2 a k + b ) \mu = 0 \\ (2ak + b) \lambda + 2a \mu = p \end{array} \right. \qquad \Rightarrow \qquad \left\{ \begin{array}{l} \mu = 0 \\ \lambda = \frac{p}{ 2ak + b } \end{array} \right. \end{align} $$ This proves the absence of the $ x^2 \textrm e^{ kx } $ terms and indicates coupling between the coefficients $\lambda$ and $\mu$.

 

(iii) Step 3. We investigate a little further the coupling between the coefficients of $x \textrm e^{kx}$, $x^2 \textrm e^{kx}$ and $x^3 \textrm e^{kx}$ terms. Consider $$ \begin{align} a \frac{ \textrm{d}^2y }{ \textrm dx^2 } + b \frac{ \textrm dy }{ \textrm dx } + c y = ( p + q x ) \textrm e^{ k x } \end{align} $$ and $$ \begin{align} y &= \nu x^3 \textrm e^{ k x } \\ y' &= 3 \nu x^2 \textrm e^{ k x } + k \nu x^3 \textrm e^{ k x } \\ &= \left( 3 x^2 + k x^3 \right) \nu \textrm e^{ k x } \\ y'' &= \left( 6 x + 3 k x^2 \right) \nu \textrm e^{ k x } + k \left( 3 x^2 + k x^3 \right) \nu \textrm e^{ k x } \\ &= \left( 6 x + 6 k x^2 + k^2 x^3 \right) \nu \textrm e^{ k x } \end{align} $$

 

We combine this with the results from $x \textrm e^{ k x }$ and $x^2 \textrm e^{ k x }$ terms, $$ \begin{align} y &= \left( \lambda x + \mu x^2 + \nu x^3 \right) \textrm e^{ k x } \\ y' &= ( 1 + k x ) \lambda \textrm e^{ k x } + \left( 2 x + k x^2 \right) \mu \textrm e^{ k x } + \left( 3 x^2 + k x^3 \right) \nu \textrm e^{ k x } \\ &= \Big[ \lambda + ( k \lambda + 2 \mu ) x + ( k \mu + 3 \nu ) x^2 + k \nu x^3 \Big] \textrm e^{ k x } \\ y'' &= \left( 2k + k^2 x \right) \lambda \textrm e^{ k x } + \left( 2 + 4 k x + k^2 x^2 \right) \mu \textrm e^{ k x } + \left( 6 x + 6 k x^2 + k^2 x^3 \right) \nu \textrm e^{ k x } \\ &= \Big[ ( 2 k \lambda + 2 \mu ) + \left( k^2 \lambda + 4 k \mu + 6 \nu \right) x + \left( k^2 \mu + 6 k \nu \right) x^2 + k^2 \nu x^3 \Big] \textrm e^{ k x } \end{align} $$ Substituting these into the differential equation gives $$ \begin{align} a y'' + b y' + c y &= a \Big[ ( 2 k \lambda + 2 \mu ) + \left( k^2 \lambda + 4 k \mu + 6 \nu \right) x + \left( k^2 \mu + 6 k \nu \right) x^2 + k^2 \nu x^3 \Big] \textrm e^{ k x } \\ &\qquad + b \Big[ \lambda + ( k \lambda + 2 \mu ) x + ( k \mu + 3 \nu ) x^2 + k \nu x^3 \Big] \textrm e^{ k x } \\ &\qquad + c \Big[ \lambda x + \mu x^2 + \nu x^3 \Big] \textrm e^{ k x } \\ &= \Big[ (2ak + b) \lambda + 2a \mu + \underbrace{ \left( a k^2 + b k + c \right) }_{ = 0 } \lambda x + 2 ( 2 a k + b ) \mu x + 6 a \nu x \\ &\qquad + \underbrace{ \left( a k^2 + b k + c \right) }_{ = 0 } \mu x^2 + 3 ( 2 a k + b ) \nu x^2 + \underbrace{ \left( a k^2 + b k + c \right) }_{ = 0 } \nu x^3 \Big] \textrm e^{ k x } \\ &= \Big[ (2ak + b) \lambda + 2a \mu + 2 ( 2 a k + b ) \mu x + 6 a \nu x + 3 ( 2 a k + b ) \nu x^2 \Big] \textrm e^{ k x } \\ &\stackrel{!}{=} ( p + qx ) \textrm e^{ k x } \end{align} $$ i.e. $$ \begin{align} \left\{ \begin{array}{l} ( 2 a k + b ) \nu = 0 \\ 2 ( 2 a k + b ) \mu + 6 a \nu = q \\ (2ak + b) \lambda + 2a \mu = p \end{array} \right. \qquad \Rightarrow \qquad \left\{ \begin{array}{l} \nu = 0 \\ \mu = \frac{ q }{ 2 ( 2 a k + b ) } \\ \lambda = \frac{ p - 2 a \mu }{ 2ak + b } = \frac{ p - 2 a \left( \frac{ q }{ 2 ( 2 a k + b ) } \right) }{ 2ak + b } = \frac{ 2 (2 a k + b ) p - 2 a q }{ 2 ( 2 a k + b )^2 } \end{array} \right. \end{align} $$ so we observe a mixing of $p$ and $q$ in $\lambda$, i.e. $\lambda$ is not a function of $p$ only, while $\mu$ is a function of $q$ only.

 

(iv) Step 4. More generally, consider $y = \textrm A(x) \textrm e^{ kx }$ for $$ \begin{align} a \frac{ \textrm{d}^2y }{ \textrm dx^2 } + b \frac{ \textrm dy }{ \textrm dx } + c y = \textrm p(x) \textrm e^{ k x } = \left( p_0 + p_1 x + p_2 x^2 + \cdots + p_r x^r \right) \textrm e^{ k x } \end{align} $$

and

$$ \begin{align} y &= \textrm A(x) \textrm e^{ k x } = \sum_{n=0}^N A_n x^n \textrm e^{ k x } \\ y' &= \sum_{n=1}^N n A_n x^{n-1} \textrm e^{ k x } + \sum_{n=0}^N k A_n x^n \textrm e^{ k x } \\ &= \sum_{n=0}^N \Big[ (n+1) A_{n+1} + k A_n \Big] x^n \textrm e^{ k x } \\ y'' &= \sum_{n=1}^N n \Big[ (n+1) A_{n+1} + k A_n \Big] x^{n-1} \textrm e^{ k x } + \sum_{n=0}^N k \Big[ (n+1) A_{n+1} + k A_n \Big] x^n \textrm e^{ k x } \\ &= \sum_{n=0}^N \Big[ (n+1)(n+2) A_{n+2} + k (n+1) A_{n+1} + k (n+1) A_{n+1} + k^2 A_n \Big] x^n \textrm e^{ k x } \\ &= \sum_{n=0}^N \Big[ (n+1)(n+2) A_{n+2} + 2 (n+1) k A_{n+1} + k^2 A_n \Big] x^n \textrm e^{ k x } \end{align} $$ Substituting these into the differential equation gives $$ \begin{align} a y'' + b y' + c y &= a \left( \sum_{n=0}^N \Big[ (n+1)(n+2) A_{n+2} + 2 (n+1) k A_{n+1} + k^2 A_n \Big] x^n \textrm e^{ k x } \right) \\ &\qquad + b \left( \sum_{n=0}^N \Big[ (n+1) A_{n+1} + k A_n \Big] x^n \textrm e^{ k x } \right) \\ &\qquad + c \left( \sum_{n=0}^N A_n x^n \textrm e^{ k x } \right) \\ &= \sum_{n=0}^N \Big( a (n+1)(n+2) A_{n+2} + \Big[ 2 a (n+1) k + b (n+1) \Big] A_{n+1} + \underbrace{ \Big[ a k^2 + b k + c \Big] }_{ = 0 } A_n \Big) x^n \textrm e^{ k x } \\ &= \sum_{n=0}^N \Big( a (n+1)(n+2) A_{n+2} + ( 2 a k + b ) (n+1) A_{n+1} \Big) x^n \textrm e^{ k x } \\ &\stackrel{!}{=} \left( p_0 + p_1 x + p_2 x^2 + \cdots + p_r x^r \right) \textrm e^{ k x } \end{align} $$ since $k$ is one of the roots of the auxiliary equation. We consider the first few terms: $$ \begin{align} n&=0 && \left( x^0 \textrm e^{ \alpha x } \; \textrm{terms} \right) && : & 2a A_2 + ( 2 a k + b ) A_1 &= p_0 \\ n&=1 && \left( x^1 \textrm e^{ \alpha x } \; \textrm{terms} \right) && : & 6a A_3 + 2 ( 2 a k + b ) A_2 &= p_1 \\ n&=2 && \left( x^2 \textrm e^{ \alpha x } \; \textrm{terms} \right) && : & 12a A_4 + 3 ( 2 a k + b ) A_3 &= p_2 \\ &\vdots \\ n&=r && \left( x^r \textrm e^{ \alpha x } \; \textrm{terms} \right) && : & a(r+1)(r+2) A_{r+2} + (r+1) ( 2 a k + b ) A_{r+1} &= p_r \\ n&=r+1 && \left( x^{r+1} \textrm e^{ \alpha x } \; \textrm{terms} \right) && : & a(r+2)(r+3) A_{r+3} + (r+2) ( 2 a k + b ) A_{r+2} &= 0 \\ &\vdots \\ n&=N-1 && \left( x^{N-1} \textrm e^{ \alpha x } \; \textrm{terms} \right) && : & a N (N+1) \underbrace{ A_{N+1} }_{=0} + N ( 2 a k + b ) A_{N} &= 0 \qquad \Rightarrow \qquad A_N = 0 \\ n&=N && \left( x^{N} \textrm e^{ \alpha x } \; \textrm{terms} \right) && : & a (N+1)(N+2) \underbrace{ A_{N+2} }_{=0} + (N+1) ( 2 a k + b ) \underbrace{ A_{N+1} }_{=0} &= 0 \qquad \checkmark \end{align} $$ and we find: $$ \begin{align} A_N = A_{N-1} = \cdots = A_{r+2}=0 \end{align} $$ and $$ \begin{align} A_{r+1} = \frac{ p_r }{ ( r + 1 ) ( 2 a k + b ) } \qquad \textrm{and}& \qquad a ( n + 1 ) A_{n+1} + ( 2 a k + b ) A_n = \frac{ p_{n-1} }{ n }, \quad n = 1,2,\cdots,r \\ \Rightarrow& \qquad A_n = \frac1{ 2 a k + b } \left( \frac{ p_{n-1} }{ n } - a ( n + 1 ) A_{n+1} \right) = \frac{ p_{n-1} - a n(n+1) A_{n+1} }{ n (2 a k + b) }, \quad n = 1,2,\cdots,r \end{align} $$ This shows that the particular integral is the standard form multiplied by $x$, and that the coefficients $A_n$ shows the following pattern:

• $A_{r+1}$ is a function of $p_r$ only.
• $A_{r}$ is a function of $p_r$ and $p_{r-1}$.
• $A_{r-1}$ is a function of $p_r$, $p_{r-1}$ and $p_{r-2}$.
• $A_{r-2}$ is a function of $p_r$, $p_{r-1}$, $p_{r-2}$ and $p_{r-3}$.
• $\vdots$
• $A_2$ is a function of $p_r, p_{r-1}, p_{r-2}, \cdots, p_{1}$.
• $A_1$ is a function of $p_r, p_{r-1}, p_{r-2}, \cdots, p_{0}$.
• $A_0$ is an arbitrary constant.

As a final cross-check, let $$ \begin{align} p_0 = p \qquad \textrm{and} \qquad p_r=0 \quad \textrm{for} \quad r \ge 1 \end{align} $$ we find $A_n=0$ for $n\ge2$, and for $n=1$: $$ \begin{align} A_1 = \frac{ p_{0} - 2 a A_{2} }{ 2 a k + b } = \frac{ p }{ 2 a k + b } \qquad \checkmark \end{align} $$ which agrees with the result above in (ii) Step 2.

 

Case 3. $ak^2 + bk + c = 0 $ and $b^2-4ac=0$: $$ \begin{align} a \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } + b \frac{ \textrm{d}y }{ \textrm{d}x } + c y = \textrm p(x) \textrm e^{ - \frac{b}{2a} x } \end{align} $$ where $\textrm p(x)$ is a polynomial, and the complementary function is $$ \begin{align} \textrm{C.F.} = ( A + B x ) \textrm e^{ - \frac{b}{2a} x}. \end{align} $$

Consider: (let $ k = - \frac{b}{2a} $ or equivalently $ 2 a k + b = 0$) $$ \begin{align} y &= \textrm A(x) \textrm e^{ k x } = \sum_{n=0}^\infty A_n x^n \textrm e^{ k x } \\ y' &= \sum_{n=1}^\infty n A_n x^{n-1} \textrm e^{ k x } + \sum_{n=0}^\infty k A_n x^n \textrm e^{ k x } \\ &= \sum_{n=0}^\infty \Big[ (n+1) A_{n+1} + k A_n \Big] x^n \textrm e^{ k x } \\ y'' &= \sum_{n=1}^\infty n \Big[ (n+1) A_{n+1} + k A_n \Big] x^{n-1} \textrm e^{ k x } + \sum_{n=0}^\infty k \Big[ (n+1) A_{n+1} + k A_n \Big] x^n \textrm e^{ k x } \\ &= \sum_{n=0}^\infty \Big[ (n+1)(n+2) A_{n+2} + k (n+1) A_{n+1} + k (n+1) A_{n+1} + k^2 A_n \Big] x^n \textrm e^{ k x } \\ &= \sum_{n=0}^\infty \Big[ (n+1)(n+2) A_{n+2} + 2 (n+1) k A_{n+1} + k^2 A_n \Big] x^n \textrm e^{ k x } \end{align} $$ Substituting these into the differential equation gives $$ \begin{align} a y'' + b y' + c y &= a \left( \sum_{n=0}^\infty \Big[ (n+1)(n+2) A_{n+2} + 2 (n+1) k A_{n+1} + k^2 A_n \Big] x^n \textrm e^{ k x } \right) \\ &\qquad + b \left( \sum_{n=0}^\infty \Big[ (n+1) A_{n+1} + k A_n \Big] x^n \textrm e^{ k x } \right) \\ &\qquad + c \left( \sum_{n=0}^\infty A_n x^n \textrm e^{ k x } \right) \\ &= \sum_{n=0}^\infty \Big( a (n+1)(n+2) A_{n+2} + \Big[ 2 a (n+1) k + b (n+1) \Big] A_{n+1} + \underbrace{ \Big[ a k^2 + b k + c \Big] }_{ = 0 } A_n \Big) x^n \textrm e^{ k x } \\ &= \sum_{n=0}^\infty \Big( a (n+1)(n+2) A_{n+2} + \Big[ \underbrace{ (2ak + b) }_{ = 0 } (n+1) \Big] A_{n+1} \Big) x^n \textrm e^{ k x } \\ &= \sum_{n=0}^\infty \Big[ a (n+1)(n+2) A_{n+2} \Big] x^n \textrm e^{ k x } \\ &= \textrm p(x) \textrm e^{ k x } = \left( p_0 + p_1 x + p_2 x^2 + \cdots \right) \textrm e^{ k x } \end{align} $$ since $k$ is the (repeated) root of the auxiliary equation and $ 2 a k + b = 0$. We consider the first few terms: $$ \begin{align} n=0 \; \left( x^0 \textrm e^{ \alpha x } \; \textrm{terms} \right) \; :& \qquad 2a A_2 = p_0 \qquad \Rightarrow \qquad A_2 = \frac{ p_0 }{ 2a } \\ n=1 \; \left( x^1 \textrm e^{ \alpha x } \; \textrm{terms} \right) \; :& \qquad 6a A_3 = p_1 \\ n=2 \; \left( x^2 \textrm e^{ \alpha x } \; \textrm{terms} \right) \; :& \qquad 12a A_4 = p_2 \\ \vdots& \end{align} $$ and we find: $$ \begin{align} A_{n+2} = \frac{ p_n }{ a (n+1)(n+2) } \end{align} $$

 

No constraints on $A_0$ and $A_1$, i.e. they are arbitrary constants.

 

Conclulsion: $ \textrm A(x) $ always takes the form of $ \textrm p(x)$ multiplied by $x^2$, i.e. $$ \begin{align} \textrm p(x) &= p && \Rightarrow & \textrm A(x) &= \lambda x^2 \\ \textrm p(x) &= qx && \Rightarrow & \textrm A(x) &= \mu x^3 \\ \textrm p(x) &= rx^2 && \Rightarrow & \textrm A(x) &= \nu x^4 \end{align} $$

Therefore, the general solution reads $$ \begin{align} y = \underbrace{ ( A_0 + A_1 x ) \textrm e^{ k x } }_{ \textrm{ complementary fn.} } + \underbrace{ \bar{\textrm{A}}(x) x^2 \textrm e^{ \alpha x } }_{ \textrm{ particular int.}} \end{align} $$ where $ \textrm A(x) = x^2 \bar{\textrm{A}}(x)$.

 

A useful calculation: We have dealt with an expression of the form $y = \sum A_n x^n \textrm e^{ kx }$ and it's useful to establish corresponding results for $y = \sum A_n x^n \textrm F(x)$ where $\textrm F(x)$ satisfies the homogeneous part of the differential equation, i.e. $a \textrm F'' + b \textrm F' + c \textrm F = 0$. $$ \begin{align} y &= \left( \sum_{n=0}^\infty A_n x^n \right) \textrm F(x) \\ y' &= \left( \sum_{n=1}^\infty n A_n x^{n-1} \right) \textrm F(x) + \left( \sum_{n=0}^\infty A_n x^n \right) \textrm F'(x) \\ &= \left( \sum_{n=0}^\infty (n+1) A_{n+1} x^n \right) \textrm F(x) + \left( \sum_{n=0}^\infty A_n x^n \right) \textrm F'(x) \\ y'' &= \left( \sum_{n=1}^\infty n(n+1) A_{n+1} x^{n-1} \right) \textrm F(x) + \left( \sum_{n=0}^\infty (n+1) A_{n+1} x^n \right) \textrm F'(x) + \left( \sum_{n=1}^\infty n A_n x^{n-1} \right) \textrm F'(x) + \left( \sum_{n=0}^\infty A_n x^n \right) \textrm F''(x) \\ &= \left( \sum_{n=0}^\infty (n+1)(n+2) A_{n+2} x^{n} \right) \textrm F(x) + 2 \left( \sum_{n=0}^\infty (n+1) A_{n+1} x^n \right) \textrm F'(x) + \left( \sum_{n=0}^\infty A_n x^n \right) \textrm F''(x) \end{align} $$ Substituting these into the differential equation gives $$ \begin{align} ay'' + by' + cy &= a \left[ \left( \sum_{n=0}^\infty (n+1)(n+2) A_{n+2} x^{n} \right) \textrm F(x) + 2 \left( \sum_{n=0}^\infty (n+1) A_{n+1} x^n \right) \textrm F'(x) + \left( \sum_{n=0}^\infty A_n x^n \right) \textrm F''(x) \right] \\ &\quad + b \left[ \left( \sum_{n=0}^\infty (n+1) A_{n+1} x^n \right) \textrm F(x) + \left( \sum_{n=0}^\infty A_n x^n \right) \textrm F'(x) \right] \\ &\quad + c \left( \sum_{n=0}^\infty A_n x^n \right) \textrm F(x) \\ &= \left( \sum_{n=0}^\infty \Big[ a (n+1)(n+2) A_{n+2} + b (n+1) A_{n+1} + c A_n \Big] x^{n} \right) \textrm F(x) \\ &\quad + \left( \sum_{n=0}^\infty \Big[ 2a (n+1) A_{n+1} + b A_n \Big] x^n \right) \textrm F'(x) \\ &\quad + a \left( \sum_{n=0}^\infty A_n x^n \right) \textrm F''(x) \\ &= \left( \sum_{n=0}^\infty a (n+1)(n+2) A_{n+2} x^n \right) \textrm F(x) + \left( \sum_{n=0}^\infty (n+1) A_{n+1} x^n \right) \Big[ 2a \textrm F'(x) + b \textrm F(x) \Big] \\ &\qquad + \left( \sum_{n=0}^\infty A_n x^n \right) \underbrace{ \Big[ a \textrm F''(x) + b \textrm F'(x) + c \textrm F'(x) \Big] }_{ = 0 } \\ &= \left( \sum_{n=0}^\infty \Big[ a (n+1)(n+2) A_{n+2} + b (n+1) A_{n+1} \Big] x^{n} \right) \textrm F(x) + \left( \sum_{n=0}^\infty 2a (n+1) A_{n+1} x^n \right) \textrm F'(x) \end{align} $$

 

(e) $ \textrm f(x) = p \cos \omega x + q \sin \omega x $.

Let $ y = \lambda \cos \omega x + \mu \sin \omega x $.

$$ \begin{align} y &= \lambda \cos \omega x + \mu \sin \omega x \\ y' &= - \lambda \omega \sin \omega x + \mu \omega \cos \omega x \\ y'' &= - \lambda \omega^2 \cos \omega x - \mu \omega^2 \sin \omega x \end{align} $$ Substituting these into the differential equation gives $$ \begin{align} a y'' + b y' + c y &= a \left( - \lambda \omega^2 \cos \omega x - \mu \omega^2 \sin \omega x \right) \\ &\quad + b \left( - \lambda \omega \sin \omega x + \mu \omega \cos \omega x \right) \\ &\quad + c \left( \lambda \cos \omega x + \mu \sin \omega x \right) \\ &= \left( c \lambda + b \mu \omega - a \lambda \omega^2 \right) \cos \omega x + ( c \mu - b \lambda \omega - a \mu \omega^2 ) \sin \omega x \\ &= \left[ \left( c - a \omega^2 \right) \lambda + b \omega \mu \right] \cos \omega x + \left[ \left( c - a \omega^2 \right) \mu - b \omega \lambda \right] \sin \omega x \stackrel{!}{=} p \cos \omega x + q \sin \omega x \end{align} $$ i.e. $$ \begin{align} \left\{ \begin{array}{ll} \left( c - a \omega^2 \right) \lambda + b \omega \mu = p \\ - b \omega \lambda + \left( c - a \omega^2 \right) \mu = q \end{array} \right. \qquad \Rightarrow \qquad \left\{ \begin{array}{ll} \lambda = \frac{ \left( c - a \omega^2 \right) p - ( b \omega ) q }{ \left( c - a \omega^2 \right)^2 + ( b \omega )^2 } \\ \mu = \frac{ \left( c - a \omega^2 \right) q + ( b \omega ) p }{ \left( c - a \omega^2 \right)^2 + ( b \omega )^2 } \end{array} \right. \end{align} $$ So the particular integral is $$ \begin{align} y &= \frac{ \left[ \left( c - a \omega^2 \right) p - ( b \omega ) q \right] \cos \omega x + \left[ \left( c - a \omega^2 \right) q + ( b \omega ) p \right] \sin \omega x }{ \left( c - a \omega^2 \right)^2 + ( b \omega )^2 } \\ &=\frac{ \left( c - a \omega^2 \right) ( p \cos \omega x + q \sin \omega x ) + b \omega ( p \sin \omega x - q \cos \omega x ) }{ \left( c - a \omega^2 \right)^2 + ( b \omega )^2 } \end{align} $$

 

Question 2. Consider $$ \begin{align} a \frac{ \textrm d^2y }{ \textrm dx^2 } + b \frac{ \textrm dy }{ \textrm dx } + c y = \textrm f(x) = (\textrm{Exp})\times (\textrm{Sinusoidal}). \end{align} $$ Suppose that the auxiliary equation has two complex conjugate roots, $ \alpha \pm i\omega $, i.e. the complementary function reads $$ \begin{align} y = \textrm e^{ \alpha x } ( A \cos \omega x + B \sin \omega x ) \end{align} $$ We investigate the situation where some part of $\textrm f(x)$ coincides with the C.F. and there are four cases to consider.

(i) Neither real($\alpha$) nor imaginary part($\omega$) coincides: $ \textrm f(x) = \textrm e^{ \beta x } ( p \cos kx + q \sin kx ) $

(ii) Real part($\alpha$) coincides: $ \textrm f(x) = \textrm e^{ \alpha x } ( p \cos kx + q \sin kx ) $

(iii) Imaginary part($\omega$) coincides: $ \textrm f(x) = \textrm e^{ \beta x } ( p \cos \omega x + q \sin \omega x ) $

(iv) Both real($\alpha$) and imaginary($\omega$) conicide: $ \textrm f(x) = \textrm e^{ \alpha x } ( p \cos \omega x + q \sin \omega x ) $

 

Step 1. Auxiliary equation: It is given that $ m = \alpha \pm i\omega $ satisfies the auxiliary equation, i.e. $$ \begin{align} a m^2 + bm + c &= a (\alpha \pm i \omega)^2 + b (\alpha \pm i \omega) + c \\ &= a \left( \alpha^2 - \omega^2 \pm 2 i \alpha \omega \right) + b (\alpha \pm i \omega) + c \\ &= \Big[ a \left( \alpha^2 - \omega^2 \right) + b \alpha + c \Big] \pm i ( 2 a \alpha + b ) \omega \end{align} $$ This gives

$$ \begin{align} \textrm{Re}&=0 && \Rightarrow & a \left( \alpha^2 - \omega^2 \right) + b \alpha + c &= 0 \\ \textrm{Im}&=0 && \Rightarrow & 2 a \alpha + b &= 0 \end{align} $$

 

Step 2. Particular integral:

 

(i) $ \textrm f(x) = \textrm e^{ \beta x } ( p \cos kx + q \sin kx ) $

We use the standard form for the particular integral: $$ \begin{align} y &= \textrm e^{ \beta x} ( \lambda \cos k x + \mu \sin k x ) \\ y' &= \beta \textrm e^{ \beta x} ( \lambda \cos k x + \mu \sin k x ) + \textrm e^{ \beta x} ( - \lambda k \sin k x + \mu k \cos k x ) \\ &= \textrm e^{ \beta x} \Big[ ( \beta \lambda + \mu k ) \cos k x + ( \beta \mu - \lambda k ) \sin k x \Big] \\ y'' &= \beta \textrm e^{ \beta x} \Big[ ( \beta \lambda + \mu k ) \cos k x + ( \beta \mu - \lambda k ) \sin k x \Big] + \textrm e^{ \beta x} \Big[ - ( \beta \lambda + \mu k ) k \sin k x + ( \beta \mu - \lambda k ) k \cos k x \Big] \\ &= \textrm e^{ \beta x} \Big[ \left( \beta^2 \lambda + 2 \beta \mu k - \lambda k^2 \right) \cos k x + \left( \beta^2 \mu - 2 \beta \lambda k + \mu k^2 \right) \sin k x \Big] \end{align} $$ Substituting these into the differential equation gives $$ \begin{align} a y'' + b y' + c y &= a \Big[ \left( \beta^2 \lambda + 2 \beta \mu k - \lambda k^2 \right) \cos k x + \left( \beta^2 \mu - 2 \beta \lambda k + \mu k^2 \right) \sin k x \Big] \textrm e^{ \beta x} \\ &\quad + b \Big[ ( \beta \lambda + \mu k ) \cos k x + ( \beta \mu - \lambda k ) \sin k x \Big] \textrm e^{ \beta x} \\ &\quad + c ( \lambda \cos k x + \mu \sin k x ) \textrm e^{ \beta x} \\ &= \Big[ a \left( \beta^2 \lambda + 2 \beta \mu k - \lambda k^2 \right) + b ( \beta \lambda + \mu k ) + c \lambda \Big] \textrm e^{ \beta x} \cos k x \\ &\qquad + \Big[ a \left( \beta^2 \mu - 2 \beta \lambda k + \mu k^2 \right) + b ( \beta \mu - \lambda k ) + c \mu \Big] \textrm e^{ \beta x} \sin k x \\ &= \Big[ \left( a \beta^2 + b \beta + c - a k^2 \right) \lambda + ( 2 a \beta + b ) k \mu \Big] \textrm e^{ \beta x} \cos k x \\ &\qquad + \Big[ \left( a \beta^2 + b \beta + c - a k^2 \right) \mu - ( 2 a \beta + b ) k \lambda \Big] \textrm e^{ \beta x} \sin k x \\ &\stackrel{!}{=} \textrm e^{ \beta x } ( p \cos kx + q \sin kx ) \end{align} $$

i.e.

$$ \begin{align}   & \Rightarrow &&   \left\{ \begin{array}{ll}     \left( a \beta^2 + b \beta + c - a k^2 \right) \lambda + ( 2 a \beta + b ) k \mu = p \\   - ( 2 a \beta + b ) k \lambda + \left( a \beta^2 + b \beta + c - a k^2 \right) \mu = q     \end{array} \right. \\ \\ & \Rightarrow && \left\{ \begin{array}{ll}     \lambda = \frac{ \left( a \beta^2 + b \beta + c - a k^2 \right) p - ( 2 a \beta + b ) q }{ \left( a \beta^2 + b \beta + c - a k^2 \right)^2 + ( 2 a \beta + b )^2 } \\   \mu = \frac{ ( 2 a \beta + b ) p + \left( a \beta^2 + b \beta + c - a k^2 \right) q }{ \left( a \beta^2 + b \beta + c - a k^2 \right)^2 + ( 2 a \beta + b )^2 }     \end{array} \right. \end{align} $$

and the particular integral reads $$ \begin{align} y &= \textrm e^{ \beta x } \left[ \frac{ \Big[ \left( a \beta^2 + b \beta + c - a k^2 \right) p - ( 2 a \beta + b ) q \Big] \cos kx + \Big[ ( 2 a \beta + b ) p + \left( a \beta^2 + b \beta + c - a k^2 \right) q \Big] \sin kx }{ \left( a \beta^2 + b \beta + c - a k^2 \right)^2 + ( 2 a \beta + b )^2 } \right] \\ &= \textrm e^{ \beta x } \left[ \frac{ \left( a \beta^2 + b \beta + c - a k^2 \right) ( p \cos kx + q \sin kx ) + ( 2 a \beta + b ) ( p \sin kx - q \cos kx ) }{ \left( a \beta^2 + b \beta + c - a k^2 \right)^2 + ( 2 a \beta + b )^2 } \right] \end{align} $$

 

(ii) $ \textrm f(x) = \textrm e^{ \alpha x } ( p \cos kx + q \sin kx ) $

We use the standard form for the particular integral: $$ \begin{align} y &= \textrm e^{ \alpha x} ( \lambda \cos k x + \mu \sin k x ) \\ y' &= \beta \textrm e^{ \alpha x} ( \lambda \cos k x + \mu \sin k x ) + \textrm e^{ \alpha x} ( - \lambda k \sin k x + \mu k \cos k x ) \\ &= \textrm e^{ \alpha x} \Big[ ( \beta \lambda + \mu k ) \cos k x + ( \beta \mu - \lambda k ) \sin k x \Big] \\ y'' &= \alpha \textrm e^{ \alpha x} \Big[ ( \alpha \lambda + \mu k ) \cos k x + ( \alpha \mu - \lambda k ) \sin k x \Big] + \textrm e^{ \alpha x} \Big[ - ( \alpha \lambda + \mu k ) k \sin k x + ( \alpha \mu - \lambda k ) k \cos k x \Big] \\ &= \textrm e^{ \alpha x} \Big[ \left( \alpha^2 \lambda + 2 \alpha \mu k - \lambda k^2 \right) \cos k x + \left( \alpha^2 \mu - 2 \alpha \lambda k + \mu k^2 \right) \sin k x \Big] \end{align} $$ Substituting these into the differential equation gives $$ \begin{align} a y'' + b y' + c y &= a \Big[ \left( \alpha^2 \lambda + 2 \alpha \mu k - \lambda k^2 \right) \cos k x + \left( \alpha^2 \mu - 2 \alpha \lambda k + \mu k^2 \right) \sin k x \Big] \textrm e^{ \alpha x} \\ &\quad + b \Big[ ( \alpha \lambda + \mu k ) \cos k x + ( \alpha \mu - \lambda k ) \sin k x \Big] \textrm e^{ \alpha x} \\ &\quad + c ( \lambda \cos k x + \mu \sin k x ) \textrm e^{ \alpha x} \\ &= \Big[ a \left( \alpha^2 \lambda + 2 \alpha \mu k - \lambda k^2 \right) + b ( \alpha \lambda + \mu k ) + c \lambda \Big] \textrm e^{ \alpha x} \cos k x \\ &\qquad + \Big[ a \left( \alpha^2 \mu - 2 \beta \lambda k + \mu k^2 \right) + b ( \alpha \mu - \lambda k ) + c \mu \Big] \textrm e^{ \alpha x} \sin k x \\ &= \Big[ \left( a \alpha^2 + b \alpha + c - a k^2 \right) \lambda + ( 2 a \alpha + b ) k \mu \Big] \textrm e^{ \alpha x} \cos k x \\ &\qquad + \Big[ \left( a \alpha^2 + b \alpha + c - a k^2 \right) \mu - ( 2 a \beta + b ) k \lambda \Big] \textrm e^{ \beta x} \sin k x \\ &\stackrel{!}{=} \textrm e^{ \alpha x } ( p \cos kx + q \sin kx ) \end{align} $$

i.e.

$$ \begin{align}   & \Rightarrow &&   \left\{ \begin{array}{ll}     \left( a \alpha^2 + b \alpha + c - a k^2 \right) \lambda + ( 2 a \alpha + b ) k \mu = p \\   - ( 2 a \alpha + b ) k \lambda + \left( a \alpha^2 + b \alpha + c - a k^2 \right) \mu = q     \end{array} \right.     \end{align} $$    Recall from the auxiliary equation:   $$ \begin{align}     \textrm{Re}&=0 && \Rightarrow & a \alpha^2 + b \alpha + c &= a \omega^2 \\   \textrm{Im}&=0 && \Rightarrow & 2 a \alpha + b &= 0 \end{align} $$     which gives   $$ \begin{align}   & \Rightarrow &&   \left\{ \begin{array}{ll}     a \left( \omega^2 - k^2 \right) \lambda = p \\   a \left( \omega^2 - k^2 \right) \mu = q     \end{array} \right.       \\ \\ & \Rightarrow && \left\{ \begin{array}{ll}     \lambda = \frac{ p }{ a \left( \omega^2 - k^2 \right) } \\   \mu = \frac{ q }{ a \left( \omega^2 - k^2 \right) }     \end{array} \right. \end{align} $$

and the particular integral reads $$ \begin{align} y &= \textrm e^{ \alpha x } \left[ \frac{ p }{ a \left( \omega^2 - k^2 \right) } \cos k x + \frac{ q }{ a \left( \omega^2 - k^2 \right) } \sin k x \right] \end{align} $$

 

(iii) $ \textrm f(x) = \textrm e^{ \beta x } ( p \cos \omega x + q \sin \omega x ) $

We use the standard form for the particular integral: $$ \begin{align} y &= \textrm e^{ \beta x} ( \lambda \cos \omega x + \mu \sin \omega x ) \\ y' &= \beta \textrm e^{ \beta x} ( \lambda \cos \omega x + \mu \sin \omega x ) + \textrm e^{ \beta x} ( - \lambda \omega \sin \omega x + \mu \omega \cos \omega x ) \\ &= \textrm e^{ \beta x} \Big[ ( \beta \lambda + \mu \omega ) \cos \omega x + ( \beta \mu - \lambda \omega ) \sin \omega x \Big] \\ y'' &= \beta \textrm e^{ \beta x} \Big[ ( \beta \lambda + \mu \omega ) \cos \omega x + ( \beta \mu - \lambda \omega ) \sin \omega x \Big] + \textrm e^{ \beta x} \Big[ - ( \beta \lambda + \mu \omega ) \omega \sin \omega x + ( \beta \mu - \lambda \omega ) \omega \cos \omega x \Big] \\ &= \textrm e^{ \beta x} \Big[ \left( \beta^2 \lambda + 2 \beta \mu \omega - \lambda \omega^2 \right) \cos \omega x + \left( \beta^2 \mu - 2 \beta \lambda \omega + \mu \omega^2 \right) \sin \omega x \Big] \end{align} $$ Substituting these into the differential equation gives $$ \begin{align} a y'' + b y' + c y &= a \Big[ \left( \beta^2 \lambda + 2 \beta \mu \omega - \lambda \omega^2 \right) \cos \omega x + \left( \beta^2 \mu - 2 \beta \lambda \omega + \mu \omega^2 \right) \sin \omega x \Big] \textrm e^{ \beta x} \\ &\quad + b \Big[ ( \beta \lambda + \mu \omega ) \cos \omega x + ( \beta \mu - \lambda \omega ) \sin \omega x \Big] \textrm e^{ \beta x} \\ &\quad + c ( \lambda \cos \omega x + \mu \sin \omega x ) \textrm e^{ \beta x} \\ &= \Big[ a \left( \beta^2 \lambda + 2 \beta \mu \omega - \lambda \omega^2 \right) + b ( \beta \lambda + \mu \omega ) + c \lambda \Big] \textrm e^{ \beta x} \cos \omega x \\ &\qquad + \Big[ a \left( \beta^2 \mu - 2 \beta \lambda \omega + \mu \omega^2 \right) + b ( \beta \mu - \lambda \omega ) + c \mu \Big] \textrm e^{ \beta x} \sin \omega x \\ &= \Big[ \left( a \beta^2 + b \beta + c - a \omega^2 \right) \lambda + ( 2 a \beta + b ) \omega \mu \Big] \textrm e^{ \beta x} \cos \omega x \\ &\qquad + \Big[ \left( a \beta^2 + b \beta + c - a \omega^2 \right) \mu - ( 2 a \beta + b ) \omega \lambda \Big] \textrm e^{ \beta x} \sin \omega x \\ &\stackrel{!}{=} \textrm e^{ \beta x } ( p \cos \omega x + q \sin \omega x ) \end{align} $$

i.e.

$$ \begin{align}   & \Rightarrow &&   \left\{ \begin{array}{ll}     \left( a \beta^2 + b \beta + c - a \omega^2 \right) \lambda + ( 2 a \beta + b ) \omega \mu = p \\   - ( 2 a \beta + b ) \omega \lambda + \left( a \beta^2 + b \beta + c - a \omega^2 \right) \mu = q     \end{array} \right.   \end {align} $$   Recall from the auxiliary equation:   $$ \begin{align}     \textrm{Re}&=0 && \Rightarrow & a \alpha^2 + b \alpha + c &= a \omega^2 \\   \textrm{Im}&=0 && \Rightarrow & 2 a \alpha + b &= 0 \end{align} $$     which gives   $$ \begin{align}   \left( a \beta^2 + b \beta + c \right) - a \omega^2   &= \left( a \beta^2 + b \beta + c \right) - \left( a \alpha^2 + b \alpha + c \right) \\   &= a \left( \beta^2 - \alpha^2 \right) + b ( \beta - \alpha ) \\   &= ( \beta - \alpha ) \left[ a ( \alpha + \beta ) + b \right]   \end{align} $$   and thus   $$ \begin{align}   & \Rightarrow &&   \left\{ \begin{array}{ll}     ( \beta - \alpha ) \left[ a ( \alpha + \beta ) + b \right] \lambda + ( 2 a \beta + b ) \omega \mu = p \\   - ( 2 a \beta + b ) \omega \lambda + ( \beta - \alpha ) \left[ a ( \alpha + \beta ) + b \right] \mu = q     \end{array} \right.       \\ \\ & \Rightarrow && \left\{ \begin{array}{ll}     \lambda = \frac{ ( \beta - \alpha ) \left[ a ( \alpha + \beta ) + b \right] p - ( 2 a \beta + b ) q }{ ( \beta - \alpha )^2 \left[ a ( \alpha + \beta ) + b \right]^2 + ( 2 a \beta + b )^2 } \\   \mu = \frac{ ( 2 a \beta + b ) p + ( \beta - \alpha ) \left[ a ( \alpha + \beta ) + b \right] q }{ ( \beta - \alpha )^2 \left[ a ( \alpha + \beta ) + b \right]^2 + ( 2 a \beta + b )^2 }     \end{array} \right. \end{align} $$

and the particular integral reads $$ \begin{align} y &= \textrm e^{ \beta x } \left[ \frac{ \Big( ( \beta - \alpha ) \left[ a ( \alpha + \beta ) + b \right] p - ( 2 a \beta + b ) q \Big) \cos kx + \Big( ( 2 a \beta + b ) p + ( \beta - \alpha ) \left[ a ( \alpha + \beta ) + b \right] q \Big) \sin kx }{ ( \beta - \alpha )^2 \left[ a ( \alpha + \beta ) + b \right]^2 + ( 2 a \beta + b )^2 } \right] \\ &= \textrm e^{ \beta x } \left[ \frac{ ( \beta - \alpha ) \left[ a ( \alpha + \beta ) + b \right] ( p \cos kx + q \sin kx ) + ( 2 a \beta + b ) ( p \sin kx - q \cos kx ) }{ ( \beta - \alpha )^2 \left[ a ( \alpha + \beta ) + b \right]^2 + ( 2 a \beta + b )^2 } \right] \end{align} $$

 

(iv) $ \textrm f(x) = \textrm e^{ \alpha x } ( p \cos \omega x + q \sin \omega x ) $

As a first attempt, we try the standard form for the particular integral: $$ \begin{align} y &= \textrm e^{ \alpha x} ( \lambda \cos \omega x + \mu \sin \omega x ) \\ y' &= \alpha \textrm e^{ \alpha x} ( \lambda \cos \omega x + \mu \sin \omega x ) + \textrm e^{ \alpha x} ( - \lambda \omega \sin \omega x + \mu \omega \cos \omega x ) \\ &= \textrm e^{ \alpha x} \Big[ ( \alpha \lambda + \mu \omega ) \cos \omega x + ( \alpha \mu - \lambda \omega ) \sin \omega x \Big] \\ y'' &= \alpha \textrm e^{ \alpha x} \Big[ ( \alpha \lambda + \mu \omega ) \cos \omega x + ( \alpha \mu - \lambda \omega ) \sin \omega x \Big] + \textrm e^{ \alpha x} \Big[ - ( \alpha \lambda + \mu \omega ) \omega \sin \omega x + ( \alpha \mu - \lambda \omega ) \omega \cos \omega x \Big] \\ &= \textrm e^{ \alpha x} \Big[ \left( \alpha^2 \lambda + 2 \alpha \mu \omega - \lambda \omega^2 \right) \cos \omega x + \left( \alpha^2 \mu - 2 \alpha \lambda \omega + \mu \omega^2 \right) \sin \omega x \Big] \end{align} $$ Substituting these into the differential equation gives $$ \begin{align} a y'' + b y' + c y &= a \Big[ \left( \alpha^2 \lambda + 2 \alpha \mu \omega - \lambda \omega^2 \right) \cos \omega x + \left( \alpha^2 \mu - 2 \alpha \lambda \omega + \mu \omega^2 \right) \sin \omega x \Big] \textrm e^{ \alpha x} \\ &\quad + b \Big[ ( \alpha \lambda + \mu \omega ) \cos \omega x + ( \alpha \mu - \lambda \omega ) \sin \omega x \Big] \textrm e^{ \alpha x} \\ &\quad + c ( \lambda \cos \omega x + \mu \sin \omega x ) \textrm e^{ \alpha x} \\ &= \Big[ a \left( \alpha^2 \lambda + 2 \alpha \mu \omega - \lambda \omega^2 \right) + b ( \alpha \lambda + \mu \omega ) + c \lambda \Big] \textrm e^{ \alpha x} \cos \omega x \\ &\quad + \Big[ a \left( \alpha^2 \mu - 2 \alpha \lambda \omega + \mu \omega^2 \right) + b ( \alpha \mu - \lambda \omega ) + c \mu \Big] \textrm e^{ \alpha x} \sin \omega x \\ &= \Big[ \underbrace{ \left( a \alpha^2 + b \alpha + c - a \omega^2 \right) }_{=0} \lambda + \underbrace{ ( 2 a \alpha + b ) }_{=0} \omega \mu \Big] \textrm e^{ \alpha x} \cos \omega x \\ &\quad + \Big[ \underbrace{ \left( a \alpha^2 + b \alpha + c - a \omega^2 \right) }_{=0} \mu - \underbrace{ ( 2 a \alpha + b ) }_{=0} \omega \lambda \Big] \textrm e^{ \alpha x} \sin \omega x \\ &= 0 \end{align} $$ In fact, the standard form coincides with the C.F. so this proves that indeed the C.F. satisfies the homogeneous part of the differential equation.

 

As a second attempt, we try the standard form multipled by $x$ as usual. Before we carry out a detailed calculation, notice that $$ \begin{align} y &= x \textrm F(x) \\ y' &= \textrm F(x) + x \textrm F'(x) \\ y'' &= \textrm F'(x) + \textrm F'(x) + x \textrm F''(x) \\ &= 2 \textrm F'(x) + x \textrm F''(x) \end{align} $$ where $\textrm F(x) = \textrm e^{ \alpha x} ( \lambda \cos \omega x + \mu \sin \omega x )$, the standard form. Substituting these into the differential equation gives $$ \begin{align} a y'' + b y' + c y &= a \Big[ 2 \textrm F'(x) + x \textrm F''(x) \Big] \\ & \quad + b \Big[ \textrm F(x) + x \textrm F'(x) \Big] \\ & \quad + c x \textrm F(x) \\ &= 2 a \textrm F'(x) + b \textrm F(x) + x \underbrace{ \left( a F'' + b F + c \right) }_{ = 0 } \\ &= 2a \textrm e^{ \alpha x} \Big[ ( \alpha \lambda + \mu \omega ) \cos \omega x + ( \alpha \mu - \lambda \omega ) \sin \omega x \Big] + b \textrm e^{ \alpha x} ( \lambda \cos \omega x + \mu \sin \omega x ) \\ &= \textrm e^{ \alpha x} \Big( \big[ \underbrace{ ( 2a \alpha + b ) }_{ = 0 } \lambda + 2a \mu \omega \big] \cos \omega x + \big[ \underbrace{ ( 2a \alpha + b ) }_{=0} \mu - 2a \lambda \omega \big] \sin \omega x \Big) \\ &= \textrm e^{ \alpha x} ( 2a \mu \omega \cos \omega x - 2a \lambda \omega \sin \omega x ) \\ &\stackrel{!}{=} \textrm e^{ \alpha x} ( p \cos \omega x + q \sin \omega x ) \end{align} $$

i.e.

$$ \begin{align}     \left\{ \begin{array}{ll}     - 2a \omega \lambda = q \\   2a \omega \mu = p \\       \end{array} \right.   && \Rightarrow &&& \left\{ \begin{array}{ll}     \lambda = - \frac{ q }{ 2a \omega } \\   \mu = \frac{ p }{ 2a \omega }     \end{array} \right.     \end{align} $$ So, the particular integral is   $$ \begin{align}   y = x \textrm F(x) = x \textrm e^{ \alpha x} \left( \frac{ - q \cos \omega x + p \sin \omega x }{ 2a \omega } \right)   \end{align} $$

 

Aside. In order to see the 'interaction' between different powers of $x$, we consider: $$ \begin{align} y &= x^n \textrm F(x) \\ y' &= n x^{n-1} \textrm F(x) + x^{n} \textrm F'(x) \\ y'' &= n(n-1) x^{n-2} \textrm F(x) + n x^{n-1} \textrm F'(x) + n x^{n-1} \textrm F'(x) + x^n \textrm F''(x) \\ &= n(n-1) x^{n-2} \textrm F(x) + 2 n x^{n-1} \textrm F'(x) + x^n \textrm F''(x) \end{align} $$

Substituting these into the differential equation gives $$ \begin{align} a y'' + b y' + c y &= a \Big[ n(n-1) x^{n-2} \textrm F(x) + 2 n x^{n-1} \textrm F'(x) + x^n \textrm F''(x) \Big] \\ & \quad + b \Big[ n x^{n-1} \textrm F(x) + x^{n} \textrm F'(x) \Big] \\ & \quad + c x^n \textrm F(x) \\ &= a n(n-1) x^{n-2} \textrm F(x) + n x^{n-1} \Big[ 2a \textrm F'(x) + b \textrm F(x) \Big] + x^n \underbrace{ \left( a F'' + b F + c \right) }_{ = 0 } \\ &= a n(n-1) x^{n-2} \textrm e^{ \alpha x} ( \lambda \cos \omega x + \mu \sin \omega x ) \\ &\quad   + n x^{n-1} \textrm e^{ \alpha x} \Big( 2a \big[ ( \alpha \lambda + \mu \omega ) \cos \omega x + ( \alpha \mu - \lambda \omega ) \sin \omega x \big] + b ( \lambda \cos \omega x + \mu \sin \omega x ) \Big) \\ &= a n(n-1) x^{n-2} \textrm e^{ \alpha x} ( \lambda \cos \omega x + \mu \sin \omega x ) \\ &\quad   + n x^{n-1} \textrm e^{ \alpha x } \Big( \big[ \underbrace{ ( 2a \alpha + b ) }_{ = 0 } \lambda + 2a \mu \omega \big] \cos \omega x + \big[ \underbrace{ ( 2a \alpha + b ) }_{=0} \mu - 2a \lambda \omega \big] \sin \omega x \Big) \\ &= a n(n-1) x^{n-2} \textrm e^{ \alpha x} ( \lambda \cos \omega x + \mu \sin \omega x ) + 2a \omega n x^{n-1} \textrm e^{ \alpha x } ( \mu \cos \omega x - \lambda \sin \omega x ) \end{align} $$ so the term $x^n \textrm F(x)$ contributes to both $ x^{n-2} \textrm F(x) $ and $ x^{n-1} \textrm F(x) $. From this, we may infer the following result for the particular integral: $$ \begin{align} \textrm f(x) &= \textrm e^{ \alpha x } \left[ \left( \sum_{n=0}^r p_n x^n \right) \cos \omega x + \left( \sum_{n=0}^r q_n x^n \right) \sin \omega x \right] \\ \\ \Rightarrow\ \qquad y &= \textrm e^{ \alpha x } \left[ \left( \sum_{n=1}^{r+1} \lambda_n x^n \right) \cos \omega x + \left( \sum_{n=1}^{r+1} \mu_n x^n \right) \sin \omega x \right] \end{align} $$ i.e. the particular integral takes the standard form multipled by $x$.

 

Proof. More explicitly, we try, with $\textrm F(x) = \textrm e^{ \alpha x} ( \lambda \cos \omega x + \mu \sin \omega x )$ $$ \begin{align} y &= \left( \sum_{n=0}^N A_n x^n \right) \textrm F(x) \end{align} $$ which gives (for a detailed derivation, see one of the Asides in CP2 §7.2 Second-order homogeneous differential equation) $$ \begin{align} ay'' + by' + cy &= \left( \sum_{n=0}^N \Big[ a (n+1)(n+2) A_{n+2} + b (n+1) A_{n+1} \Big] x^{n} \right) \textrm F(x) + \left( \sum_{n=0}^N 2a (n+1) A_{n+1} x^n \right) \textrm F'(x) \\ &= \left( \sum_{n=0}^N \Big[ a (n+1)(n+2) A_{n+2} + b (n+1) A_{n+1} \Big] x^{n} \right) \textrm e^{ \alpha x } \Big[ \lambda \cos \omega x + \mu \sin \omega x \Big] \\ &\quad + \left( \sum_{n=0}^N 2a (n+1) A_{n+1} x^n \right) \textrm e^{ \alpha x } \Big[ ( \alpha \lambda + \mu \omega ) \cos \omega x + ( \alpha \mu - \lambda \omega ) \sin \omega x \Big] \\ &= \textrm e^{ \alpha x } \left( \sum_{n=0}^N x^n \Big[ a (n+1)(n+2) A_{n+2} \lambda + \underbrace{ ( 2a \alpha + b ) }_{ = 0 } (n+1) A_{n+1} \lambda + 2a \omega (n+1) A_{n+1} \mu \Big] \cos \omega x \right) \\ &\quad + \textrm e^{ \alpha x } \left( \sum_{n=0}^N x^n \Big[ a (n+1)(n+2) A_{n+2} \mu + \underbrace{ ( 2a \alpha + b ) }_{ = 0 } (n+1) A_{n+1} \mu - 2a \omega (n+1) A_{n+1} \lambda \Big] \cos \omega x \right) \\ &= \textrm e^{ \alpha x } \left( \sum_{n=0}^N x^n \Big[ a (n+1)(n+2) A_{n+2} \lambda + 2a \omega (n+1) A_{n+1} \mu \Big] \cos \omega x \right) \\ &\quad + \textrm e^{ \alpha x } \left( \sum_{n=0}^N x^n \Big[ a (n+1)(n+2) A_{n+2} \mu - 2a \omega (n+1) A_{n+1} \lambda \Big] \cos \omega x \right) \\ &\stackrel{!}{=} \textrm e^{ \alpha x } \left[ \left( \sum_{n=0}^r p_n x^n \right) \cos \omega x + \left( \sum_{n=0}^r q_n x^n \right) \sin \omega x \right] \end{align} $$ which gives $$ \begin{align} a (n+1)(n+2) A_{n+2} \lambda + 2a \omega (n+1) A_{n+1} \mu &= p_n \\ a (n+1)(n+2) A_{n+2} \mu - 2a \omega (n+1) A_{n+1} \lambda &= q_n \end{align} $$ We see that the structure of the coefficients is not natural to match with the right-hand side. We could redefine them as: $$ \begin{align} \lambda_n &\equiv A_n \lambda \\ \mu_n &\equiv A_n \mu \end{align} $$ but this is equivalent to the following approach.

 

We alternatively try, with $\textrm F_n(x) = \textrm e^{ \alpha x} ( \lambda_n \cos \omega x + \mu_n \sin \omega x )$, $$ \begin{align} y &= \sum_{n=0}^N x^n \textrm F_n(x) \\ y' &= \sum_{n=1}^N n x^{n-1} \textrm F_n(x) + \sum_{n=0}^N x^n \textrm F_n'(x) \\ &= \sum_{n=0}^N (n+1) x^{n} \textrm F_{n+1}(x) + \sum_{n=0}^N x^n \textrm F_n'(x) \\ y'' &= \sum_{n=1}^N n(n+1) x^{n-1} \textrm F_{n+1}(x) + \sum_{n=0}^N (n+1) x^{n} \textrm F_{n+1}'(x) + \sum_{n=1}^N n x^{n-1} \textrm F_n'(x) + \sum_{n=0}^N x^n \textrm F_n''(x) \\ &= \sum_{n=0}^N (n+1)(n+2) x^{n} \textrm F_{n+2}(x) + \sum_{n=0}^N 2 (n+1) x^{n} \textrm F_{n+1}'(x) + \sum_{n=0}^N x^n \textrm F_n''(x) \end{align} $$ Substituting these into the differential equation gives $$ \begin{align} ay'' + by' + cy &= a \left[ \sum_{n=0}^N (n+1)(n+2) x^{n} \textrm F_{n+2}(x) + \sum_{n=0}^N 2 (n+1) x^{n} \textrm F_{n+1}'(x) + \sum_{n=0}^N x^n \textrm F_n''(x) \right] \\ &\quad + b \left[ \sum_{n=0}^N (n+1) x^{n} \textrm F_{n+1}(x) + \sum_{n=0}^N x^n \textrm F_n'(x) \right] \\ &\quad + c \sum_{n=0}^N x^n \textrm F_n(x) \\ &= \sum_{n=0}^N \Big[ a (n+1)(n+2) \textrm F_{n+2}(x) + (n+1) \big[ 2a \textrm F_{n+1}'(x) + b \textrm F_{n+1}(x) \big] + \underbrace{ a \textrm F_n''(x) + b \textrm F_n'(x) + x \textrm F_n(x) }_{ = 0 } \Big] x^{n} \end{align} $$ We find: $$ \begin{align} \textrm F_n(x) &= \textrm e^{ \alpha x } ( \lambda_n \cos \omega x + \mu_n \sin \omega x ) \\ \Rightarrow \qquad \textrm F_n'(x) &= \textrm e^{ \alpha x } \Big[ ( \alpha \lambda_n + \mu_n \omega ) \cos \omega x + ( \alpha \mu_n - \lambda_n \omega ) \sin \omega x \Big] \end{align} $$ and the differential equation reads $$ \begin{align} \Rightarrow \qquad ay'' + by' + cy &= \sum_{n=0}^N \Big[ a (n+1)(n+2) \textrm F_{n+2}(x) + (n+1) \big[ 2a \textrm F_{n+1}'(x) + b \textrm F_{n+1}(x) \big] \Big] x^{n} \\ &= \sum_{n=0}^N \Big[ a (n+1)(n+2) ( \lambda_{n+2} \cos \omega x + \mu_{n+2} \sin \omega x ) \\ &\qquad + (n+1) 2a ( \alpha \lambda_{n+1} + \mu_{n+1} \omega ) \cos \omega x + (n+1) 2a ( \alpha \mu_{n+1} - \lambda_{n+1} \omega ) \sin \omega x \\ &\qquad + (n+1) b \lambda_{n+1} \cos \omega x + (n+1) b \mu_{n+1} \sin \omega x \Big] x^{n} \textrm e^{ \alpha x } \\ &= \sum_{n=0}^N \bigg( (n+1) \Big[ a(n+2) \lambda_{n+2} + \underbrace{ ( 2a \alpha + b ) }_{ = 0 } \lambda_{n+1} + 2a \omega \mu_{n+1} \Big] \cos \omega x \\ &\qquad + (n+1) \Big[ a(n+2) \mu_{n+2} + \underbrace{ ( 2a \alpha + b ) }_{ = 0 } \mu_{n+1} - 2a \omega \lambda_{n+1} \Big] \sin \omega x \bigg) x^{n} \textrm e^{ \alpha x } \\ &= \sum_{n=0}^N \bigg( (n+1) \Big[ a(n+2) \lambda_{n+2} + 2a \omega \mu_{n+1} \Big] \cos \omega x \\ &\qquad + (n+1) \Big[ a(n+2) \mu_{n+2} - 2a \omega \lambda_{n+1} \Big] \sin \omega x \bigg) x^{n} \textrm e^{ \alpha x } \\ &\stackrel{!}{=} \sum_{n=0}^r \left( p_n \cos \omega x + q_n \sin \omega x \right) x^n \textrm e^{ \alpha x } \end{align} $$ i.e. $$ \begin{align} && a(n+1)(n+2) \lambda_{n+2} + 2a(n+1) \omega \mu_{n+1} &= p_n \\ && a(n+1)(n+2) \mu_{n+2} - 2a(n+1) \omega \lambda_{n+1} &= q_n \\ \\ &\Leftrightarrow & a(n+1)(n+2) \underbrace{ \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} }_{ \mathbb I_2 } \underbrace{ \begin{pmatrix} \lambda_{n+2} \\ \mu_{n+2} \end{pmatrix} }_{ \textbf L_{n+2} } + (n+1) \underbrace{ \begin{pmatrix} 0 & 2a \omega \\ -2a \omega & 0 \end{pmatrix} }_{ \mathbb M } \underbrace{ \begin{pmatrix} \lambda_{n+1} \\ \mu_{n+1} \end{pmatrix} }_{ \textbf L_{n+1} } &= \underbrace{ \begin{pmatrix} p_n \\ q_n \end{pmatrix} }_{ \textbf P_n } \\ \\ &\Leftrightarrow& a(n+1)(n+2) \mathbb I_2 \textbf L_{n+2} + (n+1)\mathbb M \textbf L_{n+1} &= \textbf P_n \end{align} $$ and $$ \begin{align} n&=0 && \left( x^0 \textrm e^{ \alpha x } \; \textrm{terms} \right) && : & 2a \mathbb I_2 \textbf L_{2} + \mathbb M \textbf L_{1} &= \textbf P_0 \\ n&=1 && \left( x^1 \textrm e^{ \alpha x } \; \textrm{terms} \right) && : & 6a \mathbb I_2 \textbf L_{3} + 2 \mathbb M \textbf L_{2} &= \textbf P_1 \\ n&=2 && \left( x^2 \textrm e^{ \alpha x } \; \textrm{terms} \right) && : & 12a \mathbb I_2 \textbf L_{4} + 3 \mathbb M \textbf L_{3} &= \textbf P_2 \\ &\vdots \\ n&=r && \left( x^r \textrm e^{ \alpha x } \; \textrm{terms} \right) && : & a(r+1)(r+2) \mathbb I_2 \textbf L_{r+2} + (r+1) \mathbb M \textbf L_{r+1} &= \textbf P_r \\ n&=r+1 && \left( x^{r+1} \textrm e^{ \alpha x } \; \textrm{terms} \right) && : & a(r+2)(r+3) \mathbb I_2 \textbf L_{r+3} + (r+2) \mathbb M \textbf L_{r+2} &= 0 \\ &\vdots \\ n&=N-1 && \left( x^{N-1} \textrm e^{ \alpha x } \; \textrm{terms} \right) && : & aN(N+1) \mathbb I_2 \underbrace{ \textbf L_{N+1} }_{=0} + N \mathbb M \textbf L_{N} &= 0 \qquad \Rightarrow \qquad \textbf L_N = 0 \\ n&=N && \left( x^{N} \textrm e^{ \alpha x } \; \textrm{terms} \right) && : & a(N+1)(N+2) \mathbb I_2 \underbrace{ \textbf L_{N+2} }_{=0} + N \mathbb M \underbrace{ \textbf L_{N+1} }_{=0} &= 0 \qquad \checkmark \end{align} $$ This gives $$ \begin{align} \textbf L_{N+2} = \textbf L_{N+1} = \textbf L_N = \textbf L_{N-1} = \cdots = \textbf L_{r+2} = 0 \end{align} $$ and $$ \begin{align} \textbf L_{r+1} = \frac{ 1 }{ r + 1 } \mathbb M^{-1} \textbf P_r \qquad \textrm{and}& \qquad a(n+1) \mathbb I_2 \textbf L_{n+1} + \mathbb M \textbf L_{n} = \frac{ 1 }{ n }\textbf P_{n-1}, \quad n = 1,2,\cdots,r \\ \Rightarrow& \qquad \textbf L_n = \mathbb M^{-1} \left( \frac{ 1 }{ n } \textbf P_{n-1} - a ( n + 1 ) \mathbb I_2 \textbf L_{n+1} \right), \quad n = 1,2,\cdots,r \end{align} $$ This shows that the particular integral is the standard form multiplied by $x$, and that the coefficients $\textbf L_n$ shows the following pattern:

• $\textbf L_{r+1}$ is a function of $\textbf P_r$ only.
• $\textbf L_{r}$ is a function of $\textbf P_r$ and $\textbf P_{r-1}$.
• $\textbf L_{r-1}$ is a function of $\textbf P_r$, $\textbf P_{r-1}$ and $\textbf P_{r-2}$.
• $\textbf L_{r-2}$ is a function of $\textbf P_r$, $\textbf P_{r-1}$, $\textbf P_{r-2}$ and $\textbf P_{r-3}$.
• $\vdots$
• $\textbf L_2$ is a function of $\textbf P_r, \textbf P_{r-1}, \textbf P_{r-2}, \cdots, \textbf P_{1}$.
• $\textbf L_1$ is a function of $\textbf P_r, \textbf P_{r-1}, \textbf P_{r-2}, \cdots, \textbf P_{0}$.
• $\textbf L_0$ is an arbitrary constant.

As a final cross-check, let $$ \begin{align} \textbf P_0 = \begin{pmatrix} p \\ q \end{pmatrix} \qquad \textrm{and} \qquad \textbf P_r=0 \quad \textrm{for} \quad r \ge 1 \end{align} $$ we find $\textbf L_n=0$ for $n\ge2$, and for $n=1$: $$ \begin{align} && \textbf L_1 &= \mathbb M^{-1} \Big( \textbf P_{0} - 2 a \mathbb I_2 \underbrace{ \textbf L_{2} }_{ = 0 } \Big) \\ \\ &\Rightarrow& \begin{pmatrix} \lambda_1 \\ \mu_1 \end{pmatrix} &= \frac1{ 2a \omega } \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} p \\ q \end{pmatrix} \\ &&&= \frac1{ 2a \omega } \begin{pmatrix} -q \\ p \end{pmatrix} \qquad \checkmark \end{align} $$ which agrees with the result in (iv).

6. Exercise 7C

Question 1. Solve each of the following differential equations, giving the general solution. $$ \begin{align} \textbf{(a)}&&& \frac{ \textrm{d}^2 y }{ \textrm dx^2 } + 6 \frac{ \textrm{d} y }{ \textrm dx } + 5 y = 10 &&& \textbf{(b)}&&& \frac{ \textrm{d}^2 y }{ \textrm dx^2 } - 8 \frac{ \textrm{d} y }{ \textrm dx } + 12 y = 36x \\ \\ \textbf{(c)}&&& \frac{ \textrm{d}^2 y }{ \textrm dx^2 } + \frac{ \textrm{d} y }{ \textrm dx } - 12 y = 12 \textrm e^{ 2x } &&& \textbf{(d)}&&& \frac{ \textrm{d}^2 y }{ \textrm dx^2 } + 2 \frac{ \textrm{d} y }{ \textrm dx } - 15 y = 5 \\ \\ \textbf{(e)}&&& \frac{ \textrm{d}^2 y }{ \textrm dx^2 } - 8 \frac{ \textrm{d} y }{ \textrm dx } + 16 y = 8x + 12 &&& \textbf{(f)}&&& \frac{ \textrm{d}^2 y }{ \textrm dx^2 } + 2 \frac{ \textrm{d} y }{ \textrm dx } + y = 25 \cos 2x \\ \\ \textbf{(g)}&&& \frac{ \textrm{d}^2 y }{ \textrm dx^2 } + 81 y = 15 \textrm e^{ 3x } &&& \textbf{(h)}&&& \frac{ \textrm{d}^2 y }{ \textrm dx^2 } + 4 y = \sin x \\ \\ \textbf{(i)}&&& \frac{ \textrm{d}^2 y }{ \textrm dx^2 } - 4 \frac{ \textrm{d} y }{ \textrm dx } + 5 y = 25 x^2 - 7 &&& \textbf{(j)}&&& \frac{ \textrm{d}^2 y }{ \textrm dx^2 } - 2 \frac{ \textrm{d} y }{ \textrm dx } + 26 y = \textrm e^x \end{align} $$

 

Solution.

 

(a)

$$ \begin{align} \frac{ \textrm{d}^2 y }{ \textrm dx^2 } + 6 \frac{ \textrm{d} y }{ \textrm dx } + 5 y = 10 \end{align} $$ (1) Complementary function: Let $ y = A \textrm e^{ mx }$ and the auxiliary equation reads $$ \begin{align} && m^2 + 6m + 5 &= 0 &\Rightarrow& ( m + 1 )( m + 5 ) &= 0 \\ &\Rightarrow& m = -1 \quad \textrm{or} \quad m&=-5 \end{align} $$ and the complementary function is $$ \begin{align} y_1 = A \textrm e^{ -x } + B \textrm e^{ -5x } \end{align} $$ (2) Particular integral: We try the standard form, $$ \begin{align} y_2 &= \lambda y_2' &= 0 \\ y_2'' &= 0 \end{align} $$ Substituting these into the differential equation gives $$ \begin{align} y_2'' + 6 y_2' + 5 y_2 &= 5 \lambda \stackrel{!}{=} 10 \qquad \Rightarrow \qquad \lambda = 2 \end{align} $$ and the particular integral is $$ \begin{align} y_2 = 2 \end{align} $$ Finally, the general solution reads $$ \begin{align} y = y_1 + y_2 = A \textrm e^{ -x } + B \textrm e^{ -5x } + 2 \qquad \checkmark \end{align} $$

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(b)

$$ \begin{align} \frac{ \textrm{d}^2 y }{ \textrm dx^2 } - 8 \frac{ \textrm{d} y }{ \textrm dx } + 12 y = 36x \end{align} $$ (1) Complementary function: Let $ y = A \textrm e^{ mx }$ and the auxiliary equation reads $$ \begin{align} && m^2 - 8m + 12 &= 0 &\Rightarrow& ( m - 2 )( m - 6 ) &= 0 \\ &\Rightarrow& m = 2 \quad \textrm{or} \quad m&=6 \end{align} $$ and the complementary function is $$ \begin{align} y_1 = A \textrm e^{ 2x } + B \textrm e^{ 6x } \end{align} $$ (2) Particular integral: We try the standard form. (Note: The right-hand side only has an $x$-term, but we still need to introduce a general linear expression, $\lambda + \mu x$, including the constant term, for the particular integral. One way to understand this is to view the right-hand side as $36x+0$, i.e. there is a constant term sitting there, it's just that its value happens to be zero, and we need to work out the corresponding values for $\lambda$ and $\mu$. If we did not need a constant term $\lambda$, the differential equation would tell us that $\lambda = 0$. Otherwise, a priori, we still need to introduce it and see what value it needs to take in order to satisfy the equation. See Question 4 and Challenge below for similar situations.) $$ \begin{align} y_2 &= \lambda + \mu x \\ y_2' &= \mu \\ y_2'' &= 0 \end{align} $$ Substituting these into the differential equation gives $$ \begin{align} y_2'' - 8y_2' + 12y_2 &= - 8 \mu + 12 ( \lambda + \mu x ) \\ &= ( 12 \lambda - 8 \mu ) + 12 \mu x \\ &\stackrel{!}{=} 36x \end{align} $$ i.e. $$ \begin{align} \left\{ \begin{array}{l} 12 \mu = 36 \\ 12 \lambda - 8 \mu = 0 \end{array} \right. \qquad \Rightarrow \qquad \left\{ \begin{array}{l} \mu = 3 \\ \lambda = \frac23 \mu = 2 \end{array} \right. \end{align} $$ and the particular integral is $$ \begin{align} y_2 &= 2 + 3x \end{align} $$ Comment: Here we see that $\lambda \ne 0$.

 

Finally, the general solution reads $$ \begin{align} y = y_1 + y_2 = A \textrm e^{ 2x } + B \textrm e^{ 6x } + 2 + 3x \qquad \checkmark \end{align} $$

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(c)

$$ \begin{align} \frac{ \textrm{d}^2 y }{ \textrm dx^2 } + \frac{ \textrm{d} y }{ \textrm dx } - 12 y = 12 \textrm e^{ 2x } \end{align} $$ (1) Complementary function: Let $ y = A \textrm e^{ mx }$ and the auxiliary equation reads $$ \begin{align} && m^2 + m - 12 &= 0 &\Rightarrow& ( m + 4 )( m - 3 ) &= 0 \\ &\Rightarrow& m = -4 \quad \textrm{or} \quad m&=3 \end{align} $$ and the complementary function is $$ \begin{align} y_1 = A \textrm e^{ -4x } + B \textrm e^{ 3x } \end{align} $$ (2) Particular integral: We try the standard form, $$ \begin{align} y_2 &= \lambda \textrm e^{ 2x } \\ y_2' &= 2 \lambda \textrm e^{ 2x } \\ y_2'' &= 4 \lambda \textrm e^{ 2x } \end{align} $$ Substituting these into the differential equation gives $$ \begin{align} y_2'' + y_2' - 12y_2 &= 4 \lambda \textrm e^{ 2x } + 2 \lambda \textrm e^{ 2x } - 12 \lambda \textrm e^{ 2x } \\ &= -6 \lambda \textrm e^{ 2x } \\ &\stackrel{!}{=} 12 \textrm e^{ 2x } \end{align} $$ We find $$ \begin{align} \lambda = -2 \end{align} $$ and the particular integral is $$ \begin{align} y_2 &= -2 \textrm e^{ 2x } \end{align} $$ Finally, the general solution reads $$ \begin{align} y = y_1 + y_2 = A \textrm e^{ -4x } + B \textrm e^{ 3x } - 2 \textrm e^{ 2x } \qquad \checkmark \end{align} $$

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(d)

$$ \begin{align} \frac{ \textrm{d}^2 y }{ \textrm dx^2 } + 2 \frac{ \textrm{d} y }{ \textrm dx } - 15 y = 5 \end{align} $$ (1) Complementary function: Let $ y = A \textrm e^{ mx }$ and the auxiliary equation reads $$ \begin{align} && m^2 + 2m - 15 &= 0 &\Rightarrow& ( m + 5 )( m - 3 ) &= 0 \\ &\Rightarrow& m = -6 \quad \textrm{or} \quad m&=3 \end{align} $$ and the complementary function is $$ \begin{align} y_1 = A \textrm e^{ -5x } + B \textrm e^{ 3x } \end{align} $$ (2) Particular integral: We try the standard form, $$ \begin{align} y_2 &= \lambda \\ y_2' &= 0 \\ y_2'' &= 0 \end{align} $$ Substituting these into the differential equation gives $$ \begin{align} y_2'' + 2y_2' - 15y_2 &= -15 \lambda \stackrel{!}{=} 5 \qquad \Rightarrow \qquad \lambda = -\frac13 \end{align} $$ and the particular integral is $$ \begin{align} y_2 &= - \frac13 \end{align} $$ Finally, the general solution reads $$ \begin{align} y = y_1 + y_2 = A \textrm e^{ -5x } + B \textrm e^{ 3x } - \frac13 \qquad \checkmark \end{align} $$

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(e)

$$ \begin{align} \frac{ \textrm{d}^2 y }{ \textrm dx^2 } - 8 \frac{ \textrm{d} y }{ \textrm dx } + 16 y = 8x + 12 \end{align} $$ (1) Complementary function: Let $ y = A \textrm e^{ mx }$ and the auxiliary equation reads $$ \begin{align} && m^2 - 8m + 16 &= 0 &\Rightarrow& ( m - 4 )^2 &= 0 \\ &\Rightarrow& m &= 4 \quad \textrm{(a repeated root)} \end{align} $$ and the complementary function is $$ \begin{align} y_1 = ( A + Bx ) \textrm e^{ 4x } \end{align} $$ (2) Particular integral: We try the standard form, $$ \begin{align} y_2 &= \lambda + \mu x \\ y_2' &= \mu \\ y_2'' &= 0 \end{align} $$ Substituting these into the differential equation gives $$ \begin{align} y_2'' - 8y_2' + 16y_2 &= - 8 \mu + 16 ( \lambda + \mu x) \\ &= ( 16 \lambda - 8 \mu ) + 16 \mu x \\ &\stackrel{!}{=} 8x + 12 \end{align} $$ i.e. $$ \begin{align} \left\{ \begin{array}{l} 16 \mu = 8 \\ 16 \lambda - 8 \mu = 12 \end{array} \right. \qquad \Rightarrow \qquad \left\{ \begin{array}{l} \mu = \frac12 \\ \lambda = \frac{ 3 + 2 \mu }{ 4 } = \frac{ 3 + 2 \left( \frac12 \right) }{ 4 } = 1 \end{array} \right. \end{align} $$ and the particular integral is $$ \begin{align} y_2 &= 1 + \frac12 x \end{align} $$ Finally, the general solution reads $$ \begin{align} y = y_1 + y_2 = ( A + Bx ) \textrm e^{ 4x } + 1 + \frac12 x \qquad \checkmark \end{align} $$

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(f)

$$ \begin{align} \frac{ \textrm{d}^2 y }{ \textrm dx^2 } + 2 \frac{ \textrm{d} y }{ \textrm dx } + y = 25 \cos 2x \end{align} $$ (1) Complementary function: Let $ y = A \textrm e^{ mx }$ and the auxiliary equation reads $$ \begin{align} && m^2 + 2m + 1 &= 0 &\Rightarrow& ( m + 1 )^2 &= 0 \\ &\Rightarrow& m &= -1 \quad \textrm{(a repeated root)} \end{align} $$ and the complementary function is $$ \begin{align} y_1 = ( A + B x ) \textrm e^{ -x } \end{align} $$ (2) Particular integral: We try the standard form, $$ \begin{align} y_2 &= \lambda \cos 2x + \mu \sin 2x \\ y_2' &= -2 \lambda \sin 2x + 2 \mu \cos 2x \\ y_2'' &= -4 \lambda \cos 2x - 4 \mu \sin 2x \end{align} $$ Substituting these into the differential equation gives $$ \begin{align} y_2'' + 2y_2' + y_2 &= ( -4 \lambda \cos 2x - 4 \mu \sin 2x ) + 2 ( -2 \lambda \sin 2x + 2 \mu \cos 2x ) + ( \lambda \cos 2x + \mu \sin 2x ) \\ &= ( -3 \lambda + 4 \mu ) \cos 2x + ( -4 \lambda - 3 \mu ) \sin 2x \\ &\stackrel{!}{=} 25 \cos 2x \end{align} $$ i.e. $$ \begin{align} \left\{ \begin{array}{l} -3 \lambda + 4 \mu = 25 \\ 4 \lambda + 3 \mu = 0 \end{array} \right. \qquad \Rightarrow \qquad \left\{ \begin{array}{l} \lambda = -3 \\ \mu = -\frac43 \lambda = 4 \end{array} \right. \end{align} $$ and the particular integral is $$ \begin{align} y_2 &= - 3 \cos 2x + 4 \sin 2x \end{align} $$ Finally, the general solution reads $$ \begin{align} y = y_1 + y_2 = ( A + B x ) \textrm e^{ -x } - 3 \cos 2x + 4 \sin 2x \qquad \checkmark \end{align} $$

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(g)

$$ \begin{align} \frac{ \textrm{d}^2 y }{ \textrm dx^2 } + 81 y = 15 \textrm e^{ 3x } \end{align} $$ (1) Complementary function: Let $ y = A \textrm e^{ mx }$ and the auxiliary equation reads $$ \begin{align} && m^2 + 81 &= 0 &\Rightarrow& m &= \pm 9i \end{align} $$ and the complementary function is $$ \begin{align} y_1 = A \cos 9x + B \sin 9x \end{align} $$ (2) Particular integral: We try the standard form, $$ \begin{align} y_2 &= \lambda \textrm e^{ 3x } \\ y_2' &= 3 \lambda \textrm e^{ 3x } \\ y_2'' &= 9 \lambda \textrm e^{ 3x } \end{align} $$ Substituting these into the differential equation gives $$ \begin{align} y_2'' + 81y_2 &= 9 \lambda \textrm e^{ 3x } + 81 \lambda \textrm e^{ 3x } \\ &= 90 \lambda \textrm e^{ 3x } \\ &\stackrel{!}{=} 15 \textrm e^{ 3x } \end{align} $$ We find: $$ \begin{align} \lambda = \frac16 \end{align} $$ and the particular integral is $$ \begin{align} y_2 &= \frac16 \textrm e^{ 3x } \end{align} $$ Finally, the general solution reads $$ \begin{align} y = y_1 + y_2 = A \cos 9x + B \sin 9x + \frac16 \textrm e^{ 3x } \qquad \checkmark \end{align} $$

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(h)

$$ \begin{align} \frac{ \textrm{d}^2 y }{ \textrm dx^2 } + 4 y = \sin x \end{align} $$ (1) Complementary function: Let $ y = A \textrm e^{ mx }$ and the auxiliary equation reads $$ \begin{align} && m^2 + 4 &= 0 &\Rightarrow& m &= \pm 2i \end{align} $$ and the complementary function is $$ \begin{align} y_1 = A \cos 2x + B \sin 2x \end{align} $$ (2) Particular integral: We try the standard form, $$ \begin{align} y_2 &= \lambda \cos x + \mu \sin x \\ y_2' &= - \lambda \sin x + \mu \cos x \\ y_2'' &= - \lambda \cos x - \mu \sin x \end{align} $$ Substituting these into the differential equation gives $$ \begin{align} y_2'' + 4y_2 &= ( - \lambda \cos x - \mu \sin x ) + 4 ( \lambda \cos x + \mu \sin x )\\ &= 3 \lambda \cos x + 3 \mu \sin x \\ &\stackrel{!}{=} \sin x \end{align} $$ We find: $$ \begin{align} \lambda = 0 \qquad \textrm{and} \qquad \mu = \frac13 \end{align} $$ and the particular integral is $$ \begin{align} y_2 &= \frac13 \sin x \end{align} $$ Finally, the general solution reads $$ \begin{align} y = y_1 + y_2 = A \cos 2x + B \sin 2x + \frac13 \sin x \qquad \checkmark \end{align} $$

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(i)

$$ \begin{align} \frac{ \textrm{d}^2 y }{ \textrm dx^2 } - 4 \frac{ \textrm{d} y }{ \textrm dx } + 5 y = 25 x^2 - 7 \end{align} $$ (1) Complementary function: Let $ y = A \textrm e^{ mx }$ and the auxiliary equation reads $$ \begin{align} && m^2 - 4m + 5 &= 0 &\Rightarrow& ( m - 2 )^2 + 1 &= 0 \\ &\Rightarrow& m &= 2 \pm i \end{align} $$ and the complementary function is $$ \begin{align} y_1 = \textrm e^{ 2x } ( A \cos x + B \sin x ) \end{align} $$ (2) Particular integral: We try the standard form, $$ \begin{align} y_2 &= \lambda + \mu x + \nu x^2 \\ y_2' &= \mu + 2 \nu x y_2'' &= 2 \nu \end{align} $$ Substituting these into the differential equation gives $$ \begin{align} y_2'' - 4y_2' + 5y_2 &= 2 \nu - 4 ( \mu + 2\nu x ) + 5 \left( \lambda + \mu x + \nu x^2 \right) \\ &= ( 2 \nu - 4 \mu + 5 \lambda ) + ( -8 \nu + 5 \mu ) x + 5 \nu x^2 \\ &\stackrel{!}{=} 25x^2 - 7 \end{align} $$ i.e. $$ \begin{align} \left\{ \begin{array}{l} 5 \nu = 25 \\ -8 \nu + 5 \mu = 0 \\ 2 \nu - 4 \mu + 5 \lambda = -7 \end{array} \right. \qquad \Rightarrow \qquad \left\{ \begin{array}{l} \nu = 5 \\ \mu = \frac{8}{5}\nu = 8 \\ \lambda = \frac{ -7 - 2 \nu + 4 \mu }{ 5 } = \frac{ -7 - 10 + 32 }{ 5 } = 3 \end{array} \right. \end{align} $$ and the particular integral is $$ \begin{align} y_2 &= 3 + 8 x + 5 x^2 \end{align} $$ Finally, the general solution reads $$ \begin{align} y = y_1 + y_2 = \textrm e^{ 2x } ( A \cos x + B \sin x ) + 3 + 8 x + 5 x^2 \qquad \checkmark \end{align} $$

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(j)

$$ \begin{align} \frac{ \textrm{d}^2 y }{ \textrm dx^2 } - 2 \frac{ \textrm{d} y }{ \textrm dx } + 26 y = \textrm e^x \end{align} $$ (1) Complementary function: Let $ y = A \textrm e^{ mx }$ and the auxiliary equation reads $$ \begin{align} && m^2 - 2m + 26 &= 0 &\Rightarrow& ( m - 1 )^2 + 25 &= 0 \\ &\Rightarrow& m &= 1 \pm 5i \end{align} $$ and the complementary function is $$ \begin{align} y_1 = \textrm e^{ x } ( A \cos 5x + B \sin 5x ) \end{align} $$ (2) Particular integral: We try the standard form, $$ \begin{align} y_2 &= \lambda \textrm e^{ x } \\ y_2' &= \lambda \textrm e^{ x } \\ y_2'' &= \lambda \textrm e^{ x } \end{align} $$ Substituting these into the differential equation gives $$ \begin{align} y_2'' - 2y_2' + 26y_2 &= ( 1 - 2 + 26 ) \lambda \textrm e^{ x } \\ &= 25 \lambda \textrm e^{ x } \\ &\stackrel{!}{=} \textrm e^{ x } \end{align} $$ We find: $$ \begin{align} \lambda = \frac1{25} \end{align} $$ and the particular integral is $$ \begin{align} y_2 &= \frac1{25} \textrm e^{ x } \end{align} $$ Finally, the general solution reads $$ \begin{align} y = y_1 + y_2 = \textrm e^{ x } ( A \cos 5x + B \sin 5x ) + \frac1{25} \textrm e^{ x } \qquad \checkmark \end{align} $$

 

Question 2. (E)

(a) Find a particular integral for the differential equation [6 marks] $$ \begin{align} \frac{ \textrm{d}^2 y }{ \textrm dx^2 } - 5 \frac{ \textrm{d} y }{ \textrm dx } + 4 y = x^2 - 3x + 2 \end{align} $$

 

(b) Hence find the general solution. [3 marks]

 

Solution.

Complementary function: Let $ y = A \textrm e^{ mx }$ and the auxiliary equation reads $$ \begin{align} && m^2 - 5m + 4 &= 0 \\ &\Rightarrow& ( m - 1 )( m - 4 ) &= 0 \\ &\Rightarrow& m = 1 \quad \textrm{or} \quad m &= 4 \end{align} $$ and the complementary function is $$ \begin{align} y_1 = A \textrm e^{ x } + B \textrm e^{ 4x } \end{align} $$

 

Particular integral: We try the standard form, $$ \begin{align} y_2 &= \lambda + \mu x + \nu x^2 \\ y_2' &= \mu + 2 \nu x y_2'' &= 2 \nu \end{align} $$ Substituting these into the differential equation gives $$ \begin{align} y_2'' - 5y_2' + 4y_2 &= 2 \nu - 5 ( \mu + 2\nu x ) + 4 \left( \lambda + \mu x + \nu x^2 \right) \\ &= ( 2 \nu - 5 \mu + 4 \lambda ) + ( - 10 \nu + 4 \mu ) x + 4 \nu x^2 \\ &\stackrel{!}{=} x^2 - 3x + 2 \end{align} $$ i.e. $$ \begin{align} \left\{ \begin{array}{l} 4 \nu = 1 \\ - 10 \nu + 4 \mu = -3 \\ 2 \nu - 5 \mu + 4 \lambda = 2 \end{array} \right. \qquad \Rightarrow \qquad \left\{ \begin{array}{l} \nu = \frac14 \\ \mu = \frac{ -3 + 10 \nu }{ 4 } = \frac{ -3 + 10 \left( \frac14 \right) }{ 4 } = -\frac18 \\ \lambda = \frac{ 2 - 2 \nu + 5 \mu }{ 4 } = \frac{ 2 - \frac12 - \frac58 }{ 4 } = \frac{7}{32} \end{array} \right. \end{align} $$ and the particular integral is $$ \begin{align} y_2 &= \frac{7}{32} - \frac18 x + \frac14 x^2 \end{align} $$ Finally, the general solution reads $$ \begin{align} y = y_1 + y_2 = A \textrm e^{ x } + B \textrm e^{ 4x } + \frac{7}{32} - \frac18 x + \frac14 x^2 \qquad \checkmark \end{align} $$

 

Question 3. (E/P) $y$ satisfies the differential equation

$$ \begin{align} \frac{ \textrm{d}^2 y }{ \textrm dx^2 } - 6 \frac{ \textrm{d} y }{ \textrm dx } = 2x^2 - x + 1 \end{align} $$

 

(a) Find the complementary function for this differential equation. [3 marks]

 

(b) Hence find a suitable particular integral and write down the general solution to the differential equation. [7 marks]

 

[Hint: Try a particular integral of the form $ \lambda x + \mu x^2 + \nu x^3 $.]

 

Solution.

(a) Complementary function: Let $ y = A \textrm e^{ mx }$ and the auxiliary equation reads $$ \begin{align} && m^2 - 6m &= 0 \\ &\Rightarrow& m ( m - 6 ) &= 0 \\ &\Rightarrow& m = 0 \quad \textrm{or} \quad m &= 6 \end{align} $$ and, noting that $A \textrm e^{0x} = A $, the complementary function is $$ \begin{align} y_1 = A + B \textrm e^{ 6x } \end{align} $$

 

(b) Particular integral: The standard form is a quadratic expression, $\lambda + \mu x + \nu x^2$. However, since the constant term is already part of the C.F., we try the standard form multiplied by $x$, i.e. $$ \begin{align} y_2 &= \lambda x + \mu x^2 + \nu x^3 \\ y_2' &= \lambda + 2 \mu x + 3 \nu x^2 \\ y_2'' &= 2 \mu + 6 \nu x \end{align} $$ Substituting these into the differential equation gives $$ \begin{align} y_2'' - 6y_2' &= ( 2 \mu + 6 \nu x ) - 6 \left( \lambda + 2 \mu x + 3 \nu x^2 \right) \\ &= ( 2 \mu - 6 \lambda ) + ( 6 \nu - 12 \mu ) x - 18 \nu x^2 \\ &\stackrel{!}{=} 2x^2 - x + 1 \end{align} $$ i.e. $$ \begin{align} \left\{ \begin{array}{l} - 18 \nu = 2 \\ 6 \nu - 12 \mu = -1 \\ 2 \mu - 6 \lambda = 2 \end{array} \right. \qquad \Rightarrow \qquad \left\{ \begin{array}{l} \nu = - \frac19 \\ \mu = \frac{ 6 \nu + 1 }{ 12 } = \frac{ 6 \left( - \frac19 \right) + 1 }{ 12 } = \frac1{36} \\ \lambda = \frac{ 2 \mu - 2 }{ 6 } = \frac{ \frac1{18} - 2 }{ 6 } = - \frac{35}{108} \end{array} \right. \end{align} $$ and the particular integral is $$ \begin{align} y_2 &= - \frac{35}{108} x + \frac1{36} x^2 - \frac19 x^3 \end{align} $$ Finally, the general solution reads $$ \begin{align} y = y_1 + y_2 = A + B \textrm e^{ 6x } - \frac{35}{108} x + \frac1{36} x^2 - \frac19 x^3 \qquad \checkmark \end{align} $$

 

Question 4. (E/P) Find the general solution to the differential equation

$$ \begin{align} \frac{ \textrm{d}^2 y }{ \textrm dx^2 } + 4 \frac{ \textrm{d} y }{ \textrm dx } = 24 x^2 \end{align} $$ [10 marks]

 

Solution.

(i) Complementary function: Let $ y = A \textrm e^{ mx }$ and the auxiliary equation reads $$ \begin{align} && m^2 + 4m &= 0 \\ &\Rightarrow& m ( m + 4 ) &= 0 \\ &\Rightarrow& m = 0 \quad \textrm{or} \quad m &= -4 \end{align} $$ and, noting that $A \textrm e^{0x} = A $, the complementary function is $$ \begin{align} y_1 = A + B \textrm e^{ -4x } \end{align} $$

 

(ii) Particular integral: The standard form is a quadratic expression, $\lambda + \mu x + \nu x^2$. However, since the constant term is already part of the C.F., we try the standard form multiplied by $x$. (Note: The right-hand side only has an $x^2$-term, but we still need to introduce a cubic expression of the form $\lambda x + \mu x^2 + \nu x^3$ for the particular integral. One way to understand this is to view the right-hand side as $24x^2+0\cdot x + 0$, i.e. there are linear and constant terms sitting there, it's just that their values happen to be zero, and we need to work out the corresponding values for $\lambda, \mu$ and $\nu$. If we did not need any terms, the differential equation would set their coefficients to be zero. Otherwise, a priori, we still need to introduce them and see what values they need to take in order to satisfy the equation. See Question 1(b) above and Challenge below for similar situations.) $$ \begin{align} y_2 &= \lambda x + \mu x^2 + \nu x^3 \\ y_2' &= \lambda + 2 \mu x + 3 \nu x^2 \\ y_2'' &= 2 \mu + 6 \nu x \end{align} $$ Substituting these into the differential equation gives $$ \begin{align} y_2'' + 4y_2' &= ( 2 \mu + 6 \nu x ) + 4 \left( \lambda + 2 \mu x + 3 \nu x^2 \right) \\ &= ( 2 \mu + 4 \lambda ) + ( 6 \nu + 8 \mu ) x + 12 \nu x^2 \\ &\stackrel{!}{=} 24 x^2 \end{align} $$ i.e. $$ \begin{align} \left\{ \begin{array}{l} 12 \nu = 24 \\ 6 \nu + 8 \mu = 0 \\ 2 \mu + 4 \lambda = 0 \end{array} \right. \qquad \Rightarrow \qquad \left\{ \begin{array}{l} \nu = 2 \\ \mu = - \frac34 \nu = - \frac32 \\ \lambda = -\frac12 \mu = -\frac12 \left( - \frac32 \right) = \frac34 \end{array} \right. \end{align} $$ and the particular integral is $$ \begin{align} y_2 &= \frac34 x - \frac32 x^2 + 2 x^3 \end{align} $$ Comment: Here we see that $\lambda \ne 0$ and $\mu \ne 0$.

 

Finally, the general solution reads $$ \begin{align} y = y_1 + y_2 = A + B \textrm e^{ -4x } + \frac34 x - \frac32 x^2 + 2 x^3 \qquad \checkmark \end{align} $$

 

Question 5. (E/P)

(a) Explain why $ \lambda x \textrm e^{x} $ is not a suitable form for the particular integral for the differential equation [2 marks] $$ \begin{align} \frac{ \textrm{d}^2 y }{ \textrm dx^2 } - 2 \frac{ \textrm{d} y }{ \textrm dx } + y = \textrm e^x \end{align} $$

 

(b) Find the value of $\lambda$ for which $ \lambda x^2 \textrm e^{x} $ is a particular integral for the differential equation. [5 marks]

 

(c) Hence find the general solution. [3 marks]

 

Solution.

(a) Complementary function: Let $ y = A \textrm e^{ mx }$ and the auxiliary equation reads $$ \begin{align} && m^2 - 2m + 1&= 0 \\ &\Rightarrow& ( m - 1 )^2 &= 0 \\ &\Rightarrow& m &= 1 \quad \textrm{(a repeated root)} \end{align} $$ and the complementary function is $$ \begin{align} y_1 = ( A + B x ) \textrm e^x \end{align} $$ Since $ \lambda x \textrm e^x $ is already part of the C.F., if we substitute it into the differential equation, it'll yield zero. Thus, it can't play a role of the particular integral which, when substituted into the DE, gives the non-zero right-hand side.

 

Aside. We show this more explicitly: $$ \begin{align} y &= \lambda x \textrm e^x \\ y' &= \lambda \textrm e^x + \lambda x \textrm e^x \\ &= ( 1 + x ) \lambda \textrm e^x \\ y'' &= \lambda \textrm e^x + ( 1 + x ) \lambda \textrm e^x \\ &= ( 2 + x ) \lambda \textrm e^x \end{align} $$ and these give $$ \begin{align} y'' - 2y' + y &= ( 2 + x ) \lambda \textrm e^x - 2 ( 1 + x ) \lambda \textrm e^x + \lambda x \textrm e^x \\ &= ( 2 - 2 ) \lambda \textrm e^x + (1 - 2 + 1) \lambda x \textrm e^x \\ &=0 \end{align} $$

 

(b) Particular integral: The question suggests the standard form multipled by $x^2$ which is the next order, i.e. $$ \begin{align} y_2 &= \lambda x^2 \textrm e^x \\ y_2' &= 2 \lambda x \textrm e^x + \lambda x^2 \textrm e^x \\ &= \left( 2x + x^2 \right) \lambda \textrm e^x \\ y_2'' &= ( 2 + 2x ) \lambda \textrm e^x + \left( 2x + x^2 \right) \lambda \textrm e^x \\ &= \left( 2 + 4x + x^2 \right) \lambda \textrm e^x \end{align} $$ Substituting these into the differential equation gives $$ \begin{align} y_2'' - 2y_2' + y_2 &= \left( 2 + 4x + x^2 \right) \lambda \textrm e^x - 2 \left( 2x + x^2 \right) \lambda \textrm e^x + \lambda x^2 \textrm e^x \\ &= 2 \lambda \textrm e^x \\ &\stackrel{!}{=} x^2 \end{align} $$ We find $$ \begin{align} \lambda = \frac12 \end{align} $$ and the particular integral is $$ \begin{align} y_2 &= \frac12 x^2 \textrm e^x \end{align} $$ Finally, the general solution reads $$ \begin{align} y = y_1 + y_2 &= A + B x \textrm e^{ x } + \frac12 x^2 \textrm e^x \\ \left( A + B x + \frac12 x^2 \right) \textrm e^x \qquad \checkmark \end{align} $$

 

Question 6. (E/P)

$$ \begin{align} \frac{ \textrm{d}^2 y }{ \textrm dt^2 } + 4 \frac{ \textrm{d} y }{ \textrm dt } + 3 y = kt + 5 \end{align} $$ where $k$ is a constant and $ t > 0 $.

 

(a) Find the general solution to the differential equation in terms of $k$. [7 marks]

 

For large values of $t$, this general solution may be approximated by a linear function.

 

(b) Given that $ k = 6 $, find the equation of this linear function. [2 marks]

 

Solution.

(a) (i) Complementary function: Let $ y = A \textrm e^{ mt }$ and the auxiliary equation reads $$ \begin{align} && m^2 + 4m + 3 &= 0 \\ &\Rightarrow& ( m + 1 )( m + 3 ) &= 0 \\ &\Rightarrow& m = -1 \quad \textrm{or} \quad m &= -3 \end{align} $$ and the complementary function is $$ \begin{align} y_1 = A \textrm e^{ -t }+ B \textrm e^{ -3t } \end{align} $$

 

(ii) Particular integral: We try the standard form, $$ \begin{align} y_2 &= \lambda + \mu t \\ y_2' &= \mu \\ y_2'' &= 0 \end{align} $$ Substituting these into the differential equation gives $$ \begin{align} y_2'' + 4y_2' + 3y_2 &= 4 \mu + 3 ( \lambda + \mu t ) \\ &= ( 4 \mu + 3 \lambda ) + 3 \mu t \\ &\stackrel{!}{=} kt + 5 \end{align} $$ i.e. $$ \begin{align} \left\{ \begin{array}{l} 3 \mu = k \\ 4 \mu + 3 \lambda = 5 \end{array} \right. \qquad \Rightarrow \qquad \left\{ \begin{array}{l} \mu = \frac{k}{3} \\ \lambda = \frac{ 5 - 4 \mu }{ 3 } = \frac{ 5 - 4 \left( \frac{k}{3} \right) }{ 3 } = \frac{ 15 - 4k }{ 9 } \end{array} \right. \end{align} $$ and the particular integral is $$ \begin{align} y_2 &= \frac{ 15 - 4k }{ 9 } + \frac{k}{3} t \end{align} $$ Finally, the general solution reads $$ \begin{align} y = y_1 + y_2 = A \textrm e^{ -t }+ B \textrm e^{ -3t } + \frac{ 15 - 4k }{ 9 } + \frac{k}{3} t \qquad \checkmark \end{align} $$

 

(b) For large values of $t$, the complementary function with the exponential terms becomes small - an exponential decay, and the general solution is approximated by the particular integral. (For an oscillatory system, this is a common behaviour and, in this context, the complementary function is often called the transient solution and the particular integral is called the steady-state solution. See CP2 §8.3 Damped and forced harmonic motion for more details.) $$ \begin{align} y \approx \frac{ 15 - 4k }{ 9 } + \frac{k}{3} t \end{align} $$ and, for $k=6$, this becomes $$ \begin{align} y \approx 2t - 1 \end{align} $$

 

Challenge. Find the general solution of the differential equation $$ \begin{align} \frac{ \textrm{d}^2 y }{ \textrm dx^2 } + y = 5 x \textrm e^{ 2x } \end{align} $$

 

Solution.

(i) Complementary function: Let $ y = A \textrm e^{ mx }$ and the auxiliary equation reads $$ \begin{align} && m^2 + 1 &= 0 \\ &\Rightarrow& m&= \pm i \end{align} $$ and the complementary function is $$ \begin{align} y_1 = A \cos x + B \sin x \end{align} $$

 

(ii) Particular integral: We try the standard form, $(\lambda + \mu x) \textrm e^{ 2x }$. (Note: The right-hand side only has an $x \textrm e^{2x}$-term, but we still need to introduce a general linear expression of the form in front of the exponential term for the particular integral. One way to understand this is to view the right-hand side as $ ( 5x + 0 ) \textrm e^{ 2x }$, i.e. there is a constant term sitting there, it's just that its value happens to be zero, and we need to choose the corresponding values for $\lambda$ and $\mu$. If we did not need a constant term, the differential equation would tell us that $\lambda = 0$. Otherwise, a priori, we still need to introduce it and see what value it needs to take in order to satisfy the equation. Similar situations were dealt with in Question 1(b) and Question 4 above.) $$ \begin{align} y_2 &= (\lambda + \mu x) \textrm e^{ 2x } \\ y_2' &= \mu \textrm e^{ 2x } + 2 (\lambda + \mu x) \textrm e^{ 2x } \\ &= ( \mu + 2 \lambda + 2 \mu x) \textrm e^{ 2x } \\ y_2'' &= 2 \mu \textrm e^{ 2x } + 2 ( \mu + 2 \lambda + 2 \mu x) \textrm e^{ 2x } \\ &= ( 4 \mu + 4 \lambda + 4 \mu x) \textrm e^{ 2x } \\ \end{align} $$ Substituting these into the differential equation gives $$ \begin{align} y_2'' + y_2 &= ( 4 \mu + 4 \lambda + 4 \mu x) \textrm e^{ 2x } + (\lambda + \mu x) \textrm e^{ 2x } \\ &= ( 4 \mu + 5 \lambda + 5 \mu x ) \textrm e^{ 2x } \\ &\stackrel{!}{=} 5x \textrm e^{ 2x } \end{align} $$ i.e. $$ \begin{align} \left\{ \begin{array}{l} 5 \mu = 5 \\ 4 \mu + 5 \lambda = 0 \end{array} \right. \qquad \Rightarrow \qquad \left\{ \begin{array}{l} \mu = 1 \\ \lambda = - \frac45 \mu = - \frac45 \end{array} \right. \end{align} $$ and the particular integral is $$ \begin{align} y_2 &= \left( - \frac45 + x \right) \textrm e^{ 2x } \end{align} $$ Comment: Here we see that $\lambda \ne 0$.

 

Finally, the general solution reads $$ \begin{align} y = y_1 + y_2 = A \cos x + B \sin x + \left( - \frac45 + x \right) \textrm e^{ 2x } \qquad \checkmark \end{align} $$