CP2 §7.2 Second-order homogeneous differential equations

Core pure mathematics 2

Table of contents

1. First example
2. Arbitrary constants = constants of integration
3. Linear and non-linear differential equations
4. Finding the general solution with the auxiliary equation
5. More examples
6. Exercise 7B

1. First example

A second-order differential equation contains second derivatives.

 

Example 1. Find the general solution to the differential equation $$ \begin{align} \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } = 12 x \end{align} $$

 

Solution.

$$ \begin{align} && \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } &= 12 x \\ \\ &\Rightarrow& \frac{ \textrm{d} y }{ \textrm{d}x } &= \int 12 x \, \textrm{d}x \\ &&&= 6x^2 + C \\ \\ &\Rightarrow& y &= \int \left( 6x^2 + C \right) \textrm{d}x \\ &&&= 2x^3 + Cx + D \end{align} $$ where $C$ and $D$ are arbitrary constants.

2. Arbitrary constants = constants of integration

Solving differential equation is essentially to perform integration to obtain $y(x)$ from $\frac{ \textrm{d} y }{ \textrm{d}x }$ or $\frac{ \textrm{d}^2y }{ \textrm{d}x^2 }$. Solving a second-order differential equation involves integrating twice, therefore the general solution contains two constants of integration. If we wanted to find a particular solution, we would need to know two boundary conditions.

3. Linear and non-linear differential equations

(1) Linear differential equations: linear in $y$-terms including $\frac{ \textrm{d}y }{ \textrm{d}x }$ and $\frac{ \textrm{d}^2y }{ \textrm{d}x^2 }$ and so on. $$ \begin{align} \textrm{(i)}&&& \textrm a(x) \frac{ \textrm{d}y }{ \textrm{d}x } + \textrm b(x) y = \textrm Q(x) \\ \textrm{(ii)}&&& \textrm a(x) \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } + \textrm b(x) \frac{ \textrm{d}y }{ \textrm{d}x } + \textrm c(x) y = \textrm f(x) \end{align} $$

 

(2) Non-linear differential equations: $$ \begin{align} \textrm{(i)}&&& \textrm a(x) \left( \frac{ \textrm{d}y }{ \textrm{d}x } \right)^2 + \textrm b(x) y = \textrm Q(x) \\ \textrm{(ii)}&&& \textrm a(x) \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } + \textrm b(x) \left( \frac{ \textrm{d}y }{ \textrm{d}x } \right)^2 + \textrm c(x) y^3 = \textrm f(x) \end{align} $$ Here, $\left( \frac{\textrm dy}{\textrm dx} \right)^2$ and $y^3$ are the non-linear terms.

4. Finding the general solution with the auxiliary equation

Second-order linear differential equations with constant coefficients $$ \begin{align} a \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } + b \frac{ \textrm{d}y }{ \textrm{d}x } + c y &= \textrm f(x) \end{align} $$

• Homogeneous DE: $\textrm f(x)=0$.
• Non-homogeneous DE: $\textrm f(x) \ne 0$.

4.1. Auxiliary equation

Recall Exercise 7A Question 16 in CP2 §7.1 First-order differential equations. The general solution of $$ \begin{align} a \frac{ \textrm{d}y }{ \textrm{d}x } + b y = 0 \end{align} $$ is of the form $$ \begin{align} y = A \textrm e^{ kx } \end{align} $$ where $ k = - \frac{b}{a} $. We notice that $k$ is the solution to the equation $ ak + b = 0 $.

 

This suggests that an equation of the form $y = A \textrm e^{ mx }$ might also be a solution of the second-order differential equation $$ \begin{align} a \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } + b \frac{ \textrm{d}y }{ \textrm{d}x } + c y = 0 \end{align} $$ We have: $$ \begin{align} y &= A \textrm e^{ m x } \\ \frac{ \textrm{d}y }{ \textrm{d}x } &= A m \textrm e^{ m x } \\ \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } &= A m^2 \textrm e^{ m x } \end{align} $$ Substituting these into the differential equation gives $$ \begin{align} && a \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } + b \frac{ \textrm{d}y }{ \textrm{d}x } + c y &= 0 \\ &\Rightarrow& a A m^2 \textrm e^{ m x } + b A m \textrm e^{ m x } + c A \textrm e^{ m x } &= 0 \\ &\Rightarrow& \left( a m^2 + b m + c \right) A \textrm e^{ m x } &= 0 \end{align} $$ This yields the auxiliary equation: $$ \begin{align} a m^2 + b m + c = 0 \qquad \Rightarrow \qquad m_{1,2} = \frac{ -b \pm \sqrt{ b^2 - 4ac } }{ 2a } \end{align} $$ so we obtain two possible values for $m$.

 

We notice that each of $y_1 = A \textrm e^{ m_1 x }$ and $y_2 = B \textrm e^{ m_2 x }$ satisfies the differential equation and thus we may take a linear combination as the general solution. This is known as the principle of superposition. Also, see Question 7 below. $$ \begin{align} y &= y_1 + y_2 = A \textrm e^{ m_1 x } + B \textrm e^{ m_2 x } \end{align} $$ This works because the given differential operator is linear: $$ \begin{align} && \left( a \frac{ \textrm{d}^2 }{ \textrm{d}x^2 } + b \frac{ \textrm{d} }{ \textrm{d}x } + c \right) y &= 0 \\ &\Rightarrow& \left( a \frac{ \textrm{d}^2 }{ \textrm{d}x^2 } + b \frac{ \textrm{d} }{ \textrm{d}x } + c \right) (y_1 + y_2) &= 0 \\ &\Rightarrow& \underbrace{ \left( a \frac{ \textrm{d}^2y_1 }{ \textrm{d}x^2 } + b \frac{ \textrm{d}y_1 }{ \textrm{d}x } + c y_1 \right) }_{ = 0 } + \underbrace{ \left( a \frac{ \textrm{d}^2y_2 }{ \textrm{d}x^2 } + b \frac{ \textrm{d}y_2 }{ \textrm{d}x } + c y_2 \right) }_{ = 0 } &= 0 \qquad \checkmark \end{align} $$ Finally, the linear combination above carries two arbitrary constants as required.

 

An alternative approach:

An equation of the form $y = A \textrm e^{ kx }$ might also be a solution of the second-order differential equation $$ \begin{align} a \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } + b \frac{ \textrm{d}y }{ \textrm{d}x } + c y = 0 \end{align} $$ But it cannot be the general solution, as it only contains one arbitrary constant. Since two constants are necessary for a second-order differential equation, we can try a solution of the form $$ \begin{align} y = A \textrm e^{ \lambda x } + B \textrm e^{ \mu x } \end{align} $$ where $A$ and $B$ are arbitrary constants and $\lambda$ and $\mu$ are constants to be determined. $$ \begin{align} y &= A \textrm e^{ \lambda x } + B \textrm e^{ \mu x } \\ \frac{ \textrm{d}y }{ \textrm{d}x } &= A \lambda \textrm e^{ \lambda x } + B \mu \textrm e^{ \mu x } \\ \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } &= A \lambda^2 \textrm e^{ \lambda x } + B \mu^2 \textrm e^{ \mu x } \end{align} $$ Substituting these into the differential equation gives $$ \begin{align} && a \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } + b \frac{ \textrm{d}y }{ \textrm{d}x } + c y &= 0 \\ &\Rightarrow& a \left( A \lambda^2 \textrm e^{ \lambda x } + B \mu^2 \textrm e^{ \mu x } \right) + b \left( A \lambda \textrm e^{ \lambda x } + B \mu \textrm e^{ \mu x } \right) + c \left( A \textrm e^{ \lambda x } + B \textrm e^{ \mu x } \right) &= 0 \\ &\Rightarrow& a A \lambda^2 \textrm e^{ \lambda x } + a B \mu^2 \textrm e^{ \mu x } + b A \lambda \textrm e^{ \lambda x } + b B \mu \textrm e^{ \mu x } + c A \textrm e^{ \lambda x } + c B \textrm e^{ \mu x } &= 0 \\ &\Rightarrow& A \textrm e^{ \lambda x } \left( a \lambda^2 + b \lambda + c \right) + B \textrm e^{ \mu x } \left( a \mu^2 + b \mu + c \right) &= 0 \end{align} $$ This shows that the equation $y = A \textrm e^{ \lambda x } + B \textrm e^{ \mu x }$ will satisfy the original differential equation if both $\lambda$ and $\mu$ are solutions to the quadratic equation: $$ \begin{align} am^2 + bm + c = 0 \end{align} $$ This equation is called the auxiliary equation.

4.2. A classification of the general solutions

The natures of the roots of the auxiliary equation, $am^2 + bm + c = 0$, determine the general solution to the differential equation. We need to consider three different cases:

 

(i) Case 1: Two distinct real roots, $\Delta = b^2 - 4ac > 0$.

Let the two distinct real roots of the auxiliary equation be $\alpha$ and $\beta$. The general solution will be of the form $$ \begin{align} y = A \textrm e^{ \alpha x } + B \textrm e^{ \beta x } \end{align} $$ where $A$ and $B$ are arbitrary constants. (See Exercise 7B, Q1 below.)

 

(ii) Case 2: One repeated root, $\Delta = b^2 - 4ac = 0$.

Let the one repeated (real) root of the auxiliary equation be $\alpha$, i.e. $$ \begin{align} \alpha = - \frac{b}{2a} \end{align} $$ The general solution will be of the form $$ \begin{align} y = ( A + B x ) \textrm e^{ \alpha x } \end{align} $$ where $A$ and $B$ are arbitrary constants. (See Exercise 7B, Q2 below.)

 

Aside. Do we need to check any other solutions of the form $x^n \textrm e^{ \alpha x }$?

More generally, we consider $$ \begin{align} y &= \sum_{n=0}^\infty A_n x^n \textrm e^{ \alpha x } \\ y' &= \sum_{n=1}^\infty n A_n x^{n-1} \textrm e^{ \alpha x } + \sum_{n=0}^\infty \alpha A_n x^n \textrm e^{ \alpha x } \\ &= \sum_{n=0}^\infty \Big[ (n+1) A_{n+1} + \alpha A_n \Big] x^n \textrm e^{ \alpha x } \\ y'' &= \sum_{n=1}^\infty n \Big[ (n+1) A_{n+1} + \alpha A_n \Big] x^{n-1} \textrm e^{ \alpha x } + \sum_{n=0}^\infty \alpha \Big[ (n+1) A_{n+1} + \alpha A_n \Big] x^n \textrm e^{ \alpha x } \\ &= \sum_{n=0}^\infty \Big[ (n+1)(n+2) A_{n+2} + \alpha (n+1) A_{n+1} + \alpha (n+1) A_{n+1} + \alpha^2 A_n \Big] x^n \textrm e^{ \alpha x } \\ &= \sum_{n=0}^\infty \Big[ (n+1)(n+2) A_{n+2} + 2 (n+1) \alpha A_{n+1} + \alpha^2 A_n \Big] x^n \textrm e^{ \alpha x } \end{align} $$ Substituting these into the differential equation gives $$ \begin{align} a y'' + b y' + c y &= a \left( \sum_{n=0}^\infty \Big[ (n+1)(n+2) A_{n+2} + 2 (n+1) \alpha A_{n+1} + \alpha^2 A_n \Big] x^n \textrm e^{ \alpha x } \right) \\ &\qquad + b \left( \sum_{n=0}^\infty \Big[ (n+1) A_{n+1} + \alpha A_n \Big] x^n \textrm e^{ \alpha x } \right) \\ &\qquad + c \left( \sum_{n=0}^\infty A_n x^n \textrm e^{ \alpha x } \right) \\ &= \sum_{n=0}^\infty \Big( a (n+1)(n+2) A_{n+2} + \Big[ 2 a (n+1) \alpha + b (n+1) \Big] A_{n+1} + \underbrace{ \Big[ a \alpha^2 + b \alpha + c \Big] }_{ = 0 } A_n \Big) x^n \textrm e^{ \alpha x } \\ &= \sum_{n=0}^\infty \Big[ a (n+1)(n+2) A_{n+2} + \underbrace{ (2 a \alpha + b) }_{ = 0 } (n+1) A_{n+1} \Big] x^n \textrm e^{ \alpha x } \\ &= \sum_{n=0}^\infty \Big[ a (n+1)(n+2) A_{n+2} \Big] x^n \textrm e^{ \alpha x } \\ &= 0 \end{align} $$ since $\alpha$ is the root of the auxiliary equation and satisfies $ 2 a \alpha + b = 0 $. We consider the first few terms: $$ \begin{align} n=0 \; \left( x^0 \textrm e^{ \alpha x } \; \textrm{terms} \right) \; :& \qquad 2a A_2 = 0 \\ n=1 \; \left( x^1 \textrm e^{ \alpha x } \; \textrm{terms} \right) \; :& \qquad 6a A_3 = 0 \\ n=2 \; \left( x^2 \textrm e^{ \alpha x } \; \textrm{terms} \right) \; :& \qquad 12a A_4 = 0 \\ \vdots& \end{align} $$ and we find: $$ \begin{align} A_2 = A_3 = A_4 = \cdots = 0 \qquad \Leftrightarrow \qquad A_{n+2}=0, \quad n \in \mathbb Z_0^+ \end{align} $$ with no constraints on $A_0$ and $A_1$, i.e. they are arbitrary constants. Therefore, we conclude that the most general solution reads $$ \begin{align} y = ( A_0 + A_1 x ) \textrm e^{ \alpha x } \end{align} $$ as anticipated earlier.

 

(iii) Case 3: Two complex conjugate roots, $\Delta = b^2 - 4ac < 0$.

Let the two complex conjugate roots of the auxiliary equation be $p \pm qi $. The general solution will be of the form $$ \begin{align} y = \textrm e^{ p x } ( A \cos qx + B \sin qx ) \end{align} $$ where $A$ and $B$ are arbitrary constants. (See Exercise 7B, Q3 below.)

 

Aside. A slightly more detailed derivation:

The two complex conjugate roots of the auxiliary equation, $am^2+bm+c=0$, are: $$ \begin{align} \alpha,\beta = p \pm qi \end{align} $$ and it yields the general solution of the form $$ \begin{align} y &= C \textrm e^{ ( p + qi ) x } + D \textrm e^{ ( p - qi ) x } \\ &= \textrm e^{ p x } \Big[ C \textrm e^{ i qx } + D \textrm e^{ - i qx } \Big] \\ &= \textrm e^{ p x } \Big[ C ( \cos qx + i \sin qx ) + D ( \cos qx - i \sin qx ) \Big] \\ &= \textrm e^{ p x } \Big[ \underbrace{ ( C + D ) }_{ A } \cos qx + \underbrace{ i ( C - D ) }_{ B } \sin qx \Big] \\ &= \textrm e^{ p x } \Big[ A \cos qx + B \sin qx \Big] \qquad \checkmark \end{align} $$

5. More examples

Example 2. (a) Find the general solution to the equation $$ \begin{align} 2 \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } + 5 \frac{ \textrm{d}y }{ \textrm{d}x } + 3 y = 0 \end{align} $$

 

(b) Verify that your answer to part (a) satisfies the equation.

 

Solution.

(a) For $y = A \textrm e^{ m x }$, the auxiliary equation is $$ \begin{align} && 2 m^2 + 5 m + 3 &= 0 \\ &\Rightarrow& (2m+3)(m+1) &= 0 \\ &\Rightarrow& m = -\frac{3}{2} \quad \textrm{or} \quad m &= -1 \end{align} $$ and the general solution reads $$ \begin{align} y = A \textrm e^{ - \frac32 x } + B \textrm e^{ - x } \end{align} $$

 

(b) $$ \begin{align} y &= A \textrm e^{ - \frac32 x } + B \textrm e^{ - x } \\ y' &= -\frac32 A \textrm e^{ - \frac32 x } - B \textrm e^{ - x } \\ y'' &= \frac94 A \textrm e^{ - \frac32 x } + B \textrm e^{ - x } \end{align} $$ Substituting these into the differential equation gives $$ \begin{align} 2 y '' + 5 y' + 3 y &= 2 \left( \frac94 A \textrm e^{ - \frac32 x } + B \textrm e^{ - x } \right) \\ &\qquad + 5 \left( -\frac32 A \textrm e^{ - \frac32 x } - B \textrm e^{ - x } \right) \\ &\qquad + 3 \left( A \textrm e^{ - \frac32 x } + B \textrm e^{ - x } \right) \\ &= \left( \frac92 - \frac{15}{2} + 3 \right) A \textrm e^{ - \frac32 x } + \left( 2 - 5 + 3 \right) B \textrm e^{ - x } \\ &= 0 \qquad \checkmark \end{align} $$

 

Example 3. Show that $$ \begin{align} y = ( A + Bx ) \textrm e^{ 3x } \end{align} $$ satisfies the differential equation $$ \begin{align} \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } - 6 \frac{ \textrm{d}y }{ \textrm{d}x } + 9 y = 0 \end{align} $$

 

Solution.

For $y = A \textrm e^{ m x }$, the auxiliary equation is $$ \begin{align} && m^2 - 6 m + 9 &= 0 \\ &\Rightarrow& (m-3)^2 &= 0 \\ &\Rightarrow& m &= 3 \end{align} $$ and the general solution reads $$ \begin{align} y = ( A + Bx ) \textrm e^{ 3 x } \end{align} $$

 

Aside. We can verify that this solution indeed satisfies the differential equation. $$ \begin{align} y &= ( A + Bx ) \textrm e^{ 3 x } \\ \\ y' &= B \textrm e^{ 3 x } + 3 ( A + Bx ) \textrm e^{ 3 x } \\ &= ( 3A + B + 3B x ) \textrm e^{ 3 x } \\ \\ y'' &= 3B \textrm e^{ 3 x } + 3 ( 3A + B + 3B x ) \textrm e^{ 3 x } \\ &= ( 9 A + 6 B + 9 B x ) \textrm e^{ 3 x } \end{align} $$ Substituting these into the differential equation gives $$ \begin{align} y '' - 6 y' + 9 y &= \left[ ( 9 A + 6 B + 9 B x ) \textrm e^{ 3 x } \right] \\ &\qquad - 6 \left[ ( 3A + B + 3B x ) \textrm e^{ 3 x } \right] \\ &\qquad + 9 \left[ ( A + B x ) \textrm e^{ 3 x } \right] \\ &= \Big[ ( 9 A - 18 A + 9 A + 6B - 6B ) + ( 9B - 18 B + 9 B) x \Big] \textrm e^{ 3 x } \\ &= 0 \qquad \checkmark \end{align} $$

 

Example 4. Find the general solution to the differential equation $$ \begin{align} \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } + 8 \frac{ \textrm{d}y }{ \textrm{d}x } + 16 y = 0. \end{align} $$

 

Solution.

For $y = A \textrm e^{ m x }$, the auxiliary equation is $$ \begin{align} && m^2 + 8 m + 16 &= 0 \\ &\Rightarrow& (m+4)^2 &= 0 \\ &\Rightarrow& m &= - 4 \end{align} $$ and the general solution reads $$ \begin{align} y = ( A + Bx ) \textrm e^{ - 4 x } \end{align} $$

 

Example 5. Find the general solution to the differential equation $$ \begin{align} \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } - 6 \frac{ \textrm{d}y }{ \textrm{d}x } + 34 y = 0 \end{align} $$

 

Solution.

For $y = A \textrm e^{ m x }$, the auxiliary equation is $$ \begin{align} && m^2 - 6 m + 34 &= 0 \\ &\Rightarrow& (m - 3)^2 + 25 &= 0 \\ &\Rightarrow& m &= 3 \pm 5 i \end{align} $$ and the general solution reads $$ \begin{align} y = \textrm e^{ 3 x } ( A \cos 5x + B \sin 5x ) \end{align} $$

 

Example 6. Find the general solution to the differential equation $$ \begin{align} \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } + 16 y = 0 \end{align} $$

 

Solution.

For $y = A \textrm e^{ m x }$, the auxiliary equation is $$ \begin{align} && m^2 + 16 &= 0 \\ &\Rightarrow& m &= \pm 4 i \end{align} $$ and the general solution reads $$ \begin{align} y = A \cos 4x + B \sin 4x \end{align} $$

6. Exercise 7B

Question 1. Find the general solution to each of the following differential equations. $$ \begin{align} \textbf{(a)}&&& \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } + 5 \frac{ \textrm{d}y }{ \textrm{d}x } + 6 y = 0 &&& \textbf{(b)}&&& \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } - 8 \frac{ \textrm{d}y }{ \textrm{d}x } + 12 y = 0 \\ \\ \textbf{(c)}&&& \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } + 2 \frac{ \textrm{d}y }{ \textrm{d}x } - 15 y = 0 &&& \textbf{(d)}&&& \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } - 3 \frac{ \textrm{d}y }{ \textrm{d}x } - 28 y = 0 \\ \\ \textbf{(e)}&&& \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } + 5 \frac{ \textrm{d}y }{ \textrm{d}x } = 0 &&& \textbf{(f)}&&& 3 \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } + 7 \frac{ \textrm{d}y }{ \textrm{d}x } + 2 y = 0 \\ \\ \textbf{(g)}&&& 4 \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } - 7 \frac{ \textrm{d}y }{ \textrm{d}x } - 2 y = 0 &&& \textbf{(h)}&&& 15 \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } - 7 \frac{ \textrm{d}y }{ \textrm{d}x } - 2 y = 0 \end{align} $$

 

Solution.

(a) For $y = A \textrm e^{ m x }$, the auxiliary equation is $$ \begin{align} && m^2 + 5m + 6 &= 0 \\ &\Rightarrow& (m + 2)(m + 3) &= 0 \\ &\Rightarrow& m = -2 \quad \textrm{or} \quad m &= -3 \end{align} $$ and the general solution reads $$ \begin{align} y = A \textrm e^{ -2x } + B \textrm e^{ -3x } \end{align} $$

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(b) For $y = A \textrm e^{ m x }$, the auxiliary equation is $$ \begin{align} && m^2 - 8m + 12 &= 0 \\ &\Rightarrow& (m - 2)(m - 6) &= 0 \\ &\Rightarrow& m = 2 \quad \textrm{or} \quad m &= 6 \end{align} $$ and the general solution reads $$ \begin{align} y = A \textrm e^{ 2x } + B \textrm e^{ 6x } \end{align} $$

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(c) For $y = A \textrm e^{ m x }$, the auxiliary equation is $$ \begin{align} && m^2 + 2m - 15 &= 0 \\ &\Rightarrow& (m - 3)(m + 5) &= 0 \\ &\Rightarrow& m = 3 \quad \textrm{or} \quad m &= -5 \end{align} $$ and the general solution reads $$ \begin{align} y = A \textrm e^{ 3x } + B \textrm e^{ -5x } \end{align} $$

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(d) For $y = A \textrm e^{ m x }$, the auxiliary equation is $$ \begin{align} && m^2 - 3m - 28 &= 0 \\ &\Rightarrow& (m - 7)(m + 4) &= 0 \\ &\Rightarrow& m = 7 \quad \textrm{or} \quad m &= -4 \end{align} $$ and the general solution reads $$ \begin{align} y = A \textrm e^{ 7x } + B \textrm e^{ -4x } \end{align} $$

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(e) For $y = A \textrm e^{ m x }$, the auxiliary equation is $$ \begin{align} && m^2 + 5m &= 0 \\ &\Rightarrow& m(m+5) &= 0 \\ &\Rightarrow& m = 0 \quad \textrm{or} \quad m &= -5 \end{align} $$ and the general solution reads $$ \begin{align} y = A + B \textrm e^{ -5x } \end{align} $$

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(f) For $y = A \textrm e^{ m x }$, the auxiliary equation is $$ \begin{align} && 3m^2 + 7m + 2 &= 0 \\ &\Rightarrow& (3m + 1)(m + 2) &= 0 \\ &\Rightarrow& m = -\frac13 \quad \textrm{or} \quad m &= -2 \end{align} $$ and the general solution reads $$ \begin{align} y = A \textrm e^{ -\frac13 x } + B \textrm e^{ -2x } \end{align} $$

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(g) For $y = A \textrm e^{ m x }$, the auxiliary equation is $$ \begin{align} && 4m^2 - 7m - 2 &= 0 \\ &\Rightarrow& (4m + 1)(m - 2) &= 0 \\ &\Rightarrow& m = -\frac14 \quad \textrm{or} \quad m &= 2 \end{align} $$ and the general solution reads $$ \begin{align} y = A \textrm e^{ -\frac14 x } + B \textrm e^{ 2x } \end{align} $$

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(h) For $y = A \textrm e^{ m x }$, the auxiliary equation is $$ \begin{align} && 15m^2 - 7m - 2 &= 0 \\ &\Rightarrow& (5m + 1)(3m - 2) &= 0 \\ &\Rightarrow& m = -\frac15 \quad \textrm{or} \quad m &= \frac23 \end{align} $$ and the general solution reads $$ \begin{align} y = A \textrm e^{ -\frac15 x } + B \textrm e^{ \frac23 x } \end{align} $$

 

Question 2. Find the general solution to each of the following differential equations. $$ \begin{align} \textbf{(a)}&&& \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } + 10 \frac{ \textrm{d}y }{ \textrm{d}x } + 25 y = 0 &&& \textbf{(b)}&&& \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } - 18 \frac{ \textrm{d}y }{ \textrm{d}x } + 81 y = 0 \\ \\ \textbf{(c)}&&& \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } + 2 \frac{ \textrm{d}y }{ \textrm{d}x } + y = 0 &&& \textbf{(d)}&&& \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } - 8 \frac{ \textrm{d}y }{ \textrm{d}x } + 16 y = 0 \\ \\ \textbf{(e)}&&& 16 \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } + 8 \frac{ \textrm{d}y }{ \textrm{d}x } + y = 0 &&& \textbf{(f)}&&& 4 \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } - 4 \frac{ \textrm{d}y }{ \textrm{d}x } + y = 0 \\ \\ \textbf{(g)}&&& 4 \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } + 20 \frac{ \textrm{d}y }{ \textrm{d}x } + 25 y = 0 &&& \textbf{(h)}&&& \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } + 2 \sqrt{3} \frac{ \textrm{d}y }{ \textrm{d}x } + 3 y = 0 \end{align} $$

 

Solution.

(a) For $y = A \textrm e^{ m x }$, the auxiliary equation is $$ \begin{align} && m^2 + 10m + 25 &= 0 \\ &\Rightarrow& (m + 5)^2 &= 0 \\ &\Rightarrow& m &= -5 \quad (\textrm{a repeated root}) \end{align} $$ and the general solution reads $$ \begin{align} y = ( A + B x ) \textrm e^{ - 5 x } \end{align} $$

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(b) For $y = A \textrm e^{ m x }$, the auxiliary equation is $$ \begin{align} && m^2 - 18m + 81 &= 0 \\ &\Rightarrow& (m - 9)^2 &= 0 \\ &\Rightarrow& m &= 9 \quad (\textrm{a repeated root}) \end{align} $$ and the general solution reads $$ \begin{align} y = ( A + B x ) \textrm e^{ 9 x } \end{align} $$

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(c) For $y = A \textrm e^{ m x }$, the auxiliary equation is $$ \begin{align} && m^2 + 2m + 1 &= 0 \\ &\Rightarrow& (m + 1)^2 &= 0 \\ &\Rightarrow& m &= -1 \quad (\textrm{a repeated root}) \end{align} $$ and the general solution reads $$ \begin{align} y = ( A + B x ) \textrm e^{ - x } \end{align} $$

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(d) For $y = A \textrm e^{ m x }$, the auxiliary equation is $$ \begin{align} && m^2 - 8m + 16 &= 0 \\ &\Rightarrow& (m - 4)^2 &= 0 \\ &\Rightarrow& m &= 4 \quad (\textrm{a repeated root}) \end{align} $$ and the general solution reads $$ \begin{align} y = ( A + B x ) \textrm e^{ 4 x } \end{align} $$

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(e) For $y = A \textrm e^{ m x }$, the auxiliary equation is $$ \begin{align} && 16m^2 + 8m + 1 &= 0 \\ &\Rightarrow& (4m + 1)^2 &= 0 \\ &\Rightarrow& m &= -\frac14 \quad (\textrm{a repeated root}) \end{align} $$ and the general solution reads $$ \begin{align} y = ( A + B x ) \textrm e^{ - frac14 x } \end{align} $$

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(f) For $y = A \textrm e^{ m x }$, the auxiliary equation is $$ \begin{align} && 4m^2 - 4m + 1 &= 0 \\ &\Rightarrow& (2m - 1)^2 &= 0 \\ &\Rightarrow& m &= \frac12 \quad (\textrm{a repeated root}) \end{align} $$ and the general solution reads $$ \begin{align} y = ( A + B x ) \textrm e^{ \frac12 x } \end{align} $$

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(g) For $y = A \textrm e^{ m x }$, the auxiliary equation is $$ \begin{align} && 4m^2 + 20m + 25 &= 0 \\ &\Rightarrow& (2m + 5)^2 &= 0 \\ &\Rightarrow& m &= -\frac52 \quad (\textrm{a repeated root}) \end{align} $$ and the general solution reads $$ \begin{align} y = ( A + B x ) \textrm e^{ - \frac52 x } \end{align} $$

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(h) For $y = A \textrm e^{ m x }$, the auxiliary equation is $$ \begin{align} && m^2 + 2 \sqrt{3} m + 3 &= 0 \\ &\Rightarrow& \left( m + \sqrt{3} \right)^2 &= 0 \\ &\Rightarrow& m &= -\sqrt{3} \quad (\textrm{a repeated root}) \end{align} $$ and the general solution reads $$ \begin{align} y = ( A + B x ) \textrm e^{ - \sqrt{3} x } \end{align} $$

 

Question 3. Find the general solution to each of the following differential equations. $$ \begin{align} \textbf{(a)}&&& \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } + 5 \frac{ \textrm{d}y }{ \textrm{d}x } + 6 y = 0 &&& \textbf{(b)}&&& \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } - 8 \frac{ \textrm{d}y }{ \textrm{d}x } + 12 y = 0 \\ \\ \textbf{(c)}&&& \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } + 2 \frac{ \textrm{d}y }{ \textrm{d}x } - 15 y = 0 &&& \textbf{(d)}&&& \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } - 3 \frac{ \textrm{d}y }{ \textrm{d}x } - 28 y = 0 \\ \\ \textbf{(e)}&&& \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } + 5 \frac{ \textrm{d}y }{ \textrm{d}x } = 0 &&& \textbf{(f)}&&& 3 \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } + 7 \frac{ \textrm{d}y }{ \textrm{d}x } + 2 y = 0 \\ \\ \textbf{(g)}&&& 4 \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } - 7 \frac{ \textrm{d}y }{ \textrm{d}x } - 2 y = 0 &&& \textbf{(h)}&&& 15 \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } - 7 \frac{ \textrm{d}y }{ \textrm{d}x } - 2 y = 0 \end{align} $$

 

Solution.

(a) For $y = A \textrm e^{ m x }$, the auxiliary equation is $$ \begin{align} && m^2 + 25 &= 0 \\ &\Rightarrow& m &= \pm 5i \end{align} $$ and the general solution reads $$ \begin{align} y = A \cos 5x + B \sin 5x \end{align} $$

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(b) For $y = A \textrm e^{ m x }$, the auxiliary equation is $$ \begin{align} && m^2 + 81 &= 0 \\ &\Rightarrow& m &= \pm 9i \end{align} $$ and the general solution reads $$ \begin{align} y = A \cos 9x + B \sin 9x \end{align} $$

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(c) For $y = A \textrm e^{ m x }$, the auxiliary equation is $$ \begin{align} && m^2 + 1 &= 0 \\ &\Rightarrow& m &= \pm i \end{align} $$ and the general solution reads $$ \begin{align} y = A \cos x + B \sin x \end{align} $$

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(d) For $y = A \textrm e^{ m x }$, the auxiliary equation is $$ \begin{align} && 9m^2 + 16 &= 0 \\ &\Rightarrow& m &= \pm \frac{4}{3}i \end{align} $$ and the general solution reads $$ \begin{align} y = A \cos \left( \frac43 x \right) + B \sin \left( \frac43 x \right) \end{align} $$

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(e) For $y = A \textrm e^{ m x }$, the auxiliary equation is $$ \begin{align} && m^2 + 8m + 17 &= 0 \\ &\Rightarrow& (m + 4)^2 + 1 &= 0 \\ &\Rightarrow& m &= -4 \pm i \end{align} $$ and the general solution reads $$ \begin{align} y = \textrm e^{ - 4 x } ( A \cos x + B \sin x ) \end{align} $$

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(f) For $y = A \textrm e^{ m x }$, the auxiliary equation is $$ \begin{align} && m^2 - 4m + 5 &= 0 \\ &\Rightarrow& (m - 2)^2 + 1 &= 0 \\ &\Rightarrow& m &= 2 \pm i \end{align} $$ and the general solution reads $$ \begin{align} y = \textrm e^{ 2 x } ( A \cos x + B \sin x ) \end{align} $$

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(g) For $y = A \textrm e^{ m x }$, the auxiliary equation is $$ \begin{align} && m^2 + 20m + 109 &= 0 \\ &\Rightarrow& (m + 10)^2 + 9 &= 0 \\ &\Rightarrow& m &= -10 \pm 3i \end{align} $$ and the general solution reads $$ \begin{align} y = \textrm e^{ - 10 x } ( A \cos 3x + B \sin 3x ) \end{align} $$

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(h) For $y = A \textrm e^{ m x }$, the auxiliary equation is $$ \begin{align} && m^2 + \sqrt{3} m + 3 &= 0 \\ &\Rightarrow& \left( m + \frac{\sqrt{3}}{2} \right)^2 + \frac{9}{4} &= 0 \\ &\Rightarrow& m &= - \frac{\sqrt{3}}{2} \pm \frac32 i \end{align} $$ and the general solution reads $$ \begin{align} y = \textrm e^{ - \frac{\sqrt{3}}{2} x } \left[ A \cos \left( \frac32 x \right) + B \sin \left( \frac32 x \right) \right] \end{align} $$

 

Question 4. Find the general solution to each of the following differential equations. $$ \begin{align} \textbf{(a)}&&& \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } + 14 \frac{ \textrm{d}y }{ \textrm{d}x } + 49 y = 0 &&& \textbf{(b)}&&& \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } + \frac{ \textrm{d}y }{ \textrm{d}x } - 12 y = 0 \\ \\ \textbf{(c)}&&& \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } + 4 \frac{ \textrm{d}y }{ \textrm{d}x } + 13 y = 0 &&& \textbf{(d)}&&& 16 \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } - 24 \frac{ \textrm{d}y }{ \textrm{d}x } + 9 y = 0 \\ \\ \textbf{(e)}&&& 9 \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } - 6 \frac{ \textrm{d}y }{ \textrm{d}x } + 5 y = 0 &&& \textbf{(f)}&&& 6 \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } - \frac{ \textrm{d}y }{ \textrm{d}x } - 2 y = 0 \end{align} $$

 

Solution.

(a) For $y = A \textrm e^{ m x }$, the auxiliary equation is $$ \begin{align} && m^2 + 14 m + 49 &= 0 \\ &\Rightarrow& ( m + 7 )^2 &= 0 \\ &\Rightarrow& m &= - 7 \quad(\textrm{a repeated root}) \end{align} $$ and the general solution reads $$ \begin{align} y = ( A + B x ) \textrm e^{ - 7x } \end{align} $$

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(b) For $y = A \textrm e^{ m x }$, the auxiliary equation is $$ \begin{align} && m^2 + m - 12 &= 0 \\ &\Rightarrow& (m + 4)(m - 3) &= 0 \\ &\Rightarrow& m = - 4 \quad \textrm{or} \quad m &= 3 \end{align} $$ and the general solution reads $$ \begin{align} y = A \textrm e^{ - 4x } + B \textrm e^{ 3x } \end{align} $$

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(c) For $y = A \textrm e^{ m x }$, the auxiliary equation is $$ \begin{align} && m^2 + 4m + 13 &= 0 \\ &\Rightarrow& (m + 2)^2 + 9 &= 0 \\ &\Rightarrow& m &= - 2 \pm 3i \end{align} $$ and the general solution reads $$ \begin{align} y = \textrm e^{ - 2x } \left( A \cos 3x + B \sin 3x \right) \end{align} $$

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(d) For $y = A \textrm e^{ m x }$, the auxiliary equation is $$ \begin{align} && 16m^2 - 24m + 9 &= 0 \\ &\Rightarrow& (4m - 3)^2 &= 0 \\ &\Rightarrow& m &= \frac34 \quad (\textrm{a repeated root}) \end{align} $$ and the general solution reads $$ \begin{align} y = ( A + B x ) \textrm e^{ \frac34 x } \end{align} $$

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(e) For $y = A \textrm e^{ m x }$, the auxiliary equation is $$ \begin{align} 9m^2 - 6m + 5 = 0 \qquad \Rightarrow \qquad m = \frac{ 3 \pm \sqrt{ 3^2 - 45 }}{ 9 } = \frac{ 3 \pm 6i }{ 9 } = \frac13 \pm \frac23 i \end{align} $$ and the general solution reads $$ \begin{align} y = \textrm e^{ \frac13 x } \left[ A \cos \left( \frac23 x \right) + B \sin \left( \frac23 x \right) \right] \end{align} $$

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(f) For $y = A \textrm e^{ m x }$, the auxiliary equation is $$ \begin{align} && 6m^2 - m - 2 &= 0 \\ &\Rightarrow& (2m + 1)(3m - 2) &= 0 \\ &\Rightarrow& m = -\frac12 \quad \textrm{or} \quad m &= \frac23 \end{align} $$ and the general solution reads $$ \begin{align} y = A \textrm e^{ -\frac12 x } + B \textrm e^{ \frac23 x } \end{align} $$

 

Question 5. (E/P) Given the differential equation $$ \begin{align} \frac{ \textrm{d}^2x }{ \textrm{d}t^2 } + 2k \frac{ \textrm{d}x }{ \textrm{d}t } + 9 x = 0 \end{align} $$ where $k$ is a real constant,

 

(a) [8 marks] find the general solution to the differential equation in each of the following cases:

1. $ | k | > 3 $
2. $ | k | < 3 $
3. $ | k | = 3 $

(b) [4 marks] In the case where $k=2$,

1. find the general solution
2. describe what happens to $x$ as $ t \rightarrow \infty$.

 

Solution.

(a) For $x = A \textrm e^{ m t }$, the auxiliary equation is $$ \begin{align} && m^2 + 2k m + 9 &= 0 \\ &\Rightarrow& (m + k)^2 + 9 - k^2 &= 0 \\ &\Rightarrow& m &= -k \pm \sqrt{ k^2 - 9 } \end{align} $$

 

1. For $|k| > 3$, $$ \begin{align} m = -k \pm \sqrt{ k^2 - 9 } \end{align} $$ the general solution reads $$ \begin{align} x(t) &= A \textrm e^{ \left( -k + \sqrt{ k^2 - 9 } \right) t } + B \textrm e^{ \left( -k - \sqrt{ k^2 - 9 } \right) t } \\ &= \textrm e^{ - k t } \left( A \textrm e^{ \sqrt{ k^2 - 9 } t } + B \textrm e^{ - \sqrt{ k^2 - 9 } t } \right) \end{align} $$

 

2. For $|k| < 3$, $$ \begin{align} m = -k \pm i \sqrt{ 9 - k^2 } \end{align} $$ the general solution reads $$ \begin{align} x(t) = \textrm e^{ -k t } \left[ A \cos \left( \sqrt{ 9 - k^2 } t \right) + B \sin \left( \sqrt{ 9 - k^2 } t \right) \right] \end{align} $$

 

3. For $|k| = 3$, $$ \begin{align} m = -k \quad (\textrm{a repeated root}) \end{align} $$ the general solution reads $$ \begin{align} x(t) &= (A + B t) \textrm e^{ - k t } \end{align} $$

 

(b) For $k=2$, the general solution reads $$ \begin{align} x(t) = \textrm e^{ -2 t } \left[ A \cos \left( \sqrt{ 5 } t \right) + B \sin \left( \sqrt{ 5 } t \right) \right] \end{align} $$ As $t \rightarrow \infty$, $$ \begin{align} x(t) \rightarrow 0 \end{align} $$

 

Question 6. (P) Given that $ a m^2 + b m + c = 0 $ has equal roots $m = \alpha$, prove that $$ \begin{align} y = ( A + B x ) \textrm e^{ \alpha x } \end{align} $$ is a solution to the differential equation $$ \begin{align} a \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } + b \frac{ \textrm{d}y }{ \textrm{d}x } + c y = 0. \end{align} $$

 

Solution.

The repeated root $\alpha$ satisfies: $$ \begin{align} a \alpha^2 + b \alpha + c = 0 \qquad \textrm{and} \qquad 2a \alpha + b = 0 \end{align} $$ $$ \begin{align} y &= ( A + Bx ) \textrm e^{ \alpha x } \\ \\ y' &= B \textrm e^{ \alpha x } + \alpha ( A + Bx ) \textrm e^{ \alpha x } \\ &= ( \alpha A + B + \alpha B x ) \textrm e^{ \alpha x } \\ \\ y'' &= \alpha B \textrm e^{ \alpha x } + \alpha ( \alpha A + B + \alpha B x ) \textrm e^{ \alpha x } \\ &= ( \alpha^2 A + 2\alpha B + \alpha^2 B x ) \textrm e^{ \alpha x } \end{align} $$ Substituting these into the differential equation gives $$ \begin{align} a y'' + b y' + c y &= a \left[ ( \alpha^2 A + 2\alpha B + \alpha^2 B x ) \textrm e^{ \alpha x } \right] \\ &\qquad + b \left[ ( \alpha A + B + \alpha B x ) \textrm e^{ \alpha x } \right] \\ &\qquad + c ( A + Bx ) \textrm e^{ \alpha x } \\ &= \underbrace{ \left( a \alpha^2 + b \alpha + c \right) }_{=0} A \textrm e^{ \alpha x } + \underbrace{ \left( 2a \alpha + b \right) }_{=0} B \textrm e^{ \alpha x } + \underbrace{ \left( a \alpha^2 + b \alpha + c \right) }_{=0} B \textrm e^{ \alpha x } \\ &= 0 \qquad \square \end{align} $$

 

Question 7. (P) Given that $ y = \textrm f(x) $ and $ y = \textrm g(x) $ are both solutions to the second-order differential equation $$ \begin{align} a \frac{ \textrm{d}^2y }{ \textrm{d}x^2 } + b \frac{ \textrm{d}y }{ \textrm{d}x } + c y = 0, \end{align} $$ prove that $$ \begin{align} y = A \textrm f(x) + B \textrm g(x) \end{align} $$ where $A$ and $B$ are real constants, is also a solution.

 

Solution.

It is given that: $$ \begin{align} a \frac{ \textrm{d}^2 \textrm f }{ \textrm{d}x^2 } + b \frac{ \textrm{d} \textrm f }{ \textrm{d}x } + c \textrm f(x) &= 0 \\ \\ a \frac{ \textrm{d}^2 \textrm g }{ \textrm{d}x^2 } + b \frac{ \textrm{d} \textrm g }{ \textrm{d}x } + c \textrm g(x) &= 0 \\ \\ \end{align} $$ we have: $$ \begin{align} \left( a \frac{ \textrm{d}^2 }{ \textrm{d}x^2 } + b \frac{ \textrm{d}}{ \textrm{d}x } + c \right) \Big[ A \textrm f(x) + B \textrm g(x) \Big] &= A \underbrace{ \left( a \frac{ \textrm{d}^2 \textrm f }{ \textrm{d}x^2 } + b \frac{ \textrm{d} \textrm f }{ \textrm{d}x } + c \textrm f(x) \right) }_{=0} + B \underbrace{ \left( a \frac{ \textrm{d}^2 \textrm g }{ \textrm{d}x^2 } + b \frac{ \textrm{d} \textrm g }{ \textrm{d}x } + c \textrm g(x) \right) }_{=0} = 0 \qquad \square \end{align} $$

 

Challenge. Let $\alpha$ and $\beta$ be the roots of a real-valued quadratic equation, so that $$ \begin{align} \alpha = p+iq \qquad \textrm{and} \qquad \beta = p-iq, \quad p,q \in \mathbb R \end{align} $$ Show that it is possible to choose $A,B \in \mathbb C$ such that $ A \textrm e^{ \alpha x } + B \textrm e^{ \beta x } $ can be written in the form $$ \begin{align} \textrm e^{ px } ( C \cos qx + D \sin qx ) \end{align} $$ where $C$ and $D$ are arbitrary real constants.

 

Solution.

$$ \begin{align} y &= A \textrm e^{ \alpha x } + B \textrm e^{ \beta x } \\ &= A \textrm e^{ ( p + qi ) x } + B \textrm e^{ ( p - qi ) x }, \qquad A,B \in \mathbb C \\ &= \textrm e^{ p x } \Big[ A \textrm e^{ i qx } + B \textrm e^{ - i qx } \Big] \\ &= \textrm e^{ p x } \Big[ A ( \cos qx + i \sin qx ) + B ( \cos qx - i \sin qx ) \Big] \\ &= \textrm e^{ p x } \Big[ \underbrace{ ( A + B ) }_{ C } \cos qx + \underbrace{ i ( A - B ) }_{ D } \sin qx \Big] \\ &= \textrm e^{ p x } \Big[ C \cos qx + D \sin qx \Big] \qquad \checkmark \end{align} $$ with $$ \begin{align} C = A + B \qquad \textrm{and} \qquad D = i ( A - B ) \end{align} $$ For a second-order differential equation $ay''+by'+cy=0$ with $a,b,c \in \mathbb R$, we will obtain a real-valued solution, i.e. $$ \begin{align} y^*=y \qquad \Rightarrow \qquad y^* = A^* \textrm e^{ ( p - qi ) x } + B^* \textrm e^{ ( p + qi ) x } = A \textrm e^{ ( p + qi ) x } + B \textrm e^{ ( p - qi ) x } \end{align} $$ and this gives $$ \begin{align} B^* = A \end{align} $$ Let $$ \begin{align} A = \lambda + i \mu, \qquad \lambda, \mu \in \mathbb R \end{align} $$ and we find: $$ \begin{align} C &= A + B = (\lambda + i \mu) + (\lambda - i \mu) = 2 \lambda \\ D &= i ( A - B ) = i \Big[ (\lambda + i \mu) - (\lambda - i \mu) \Big] = i (2i \mu) = -2 \mu \end{align} $$ so we obtain $C,D \in \mathbb R$.

 

Aside. What if the coefficients of the differential equation are complex numbers?

$$ \begin{align} (a_1 + ia_2) \frac{ \textrm{d}^2 y }{ \textrm{d}x^2 } + (b_1 + ib_2) \frac{ \textrm{d} y }{ \textrm{d}x } + (c_1 + ic_2) y &= 0 \end{align} $$ Then we would expect to have a complex-valued solution, i.e. $$ \begin{align} y(x) = \psi(x) + i \chi(x) \end{align} $$ Substituting these into the differential equation gives $$ \begin{align} && (a_1 + ia_2) \left( \psi'' + i \chi'' \right) + (b_1 + ib_2) \left( \psi' + i \chi' \right) + (c_1 + ic_2) \left( \psi + i \chi \right) &= 0 \\ &\Rightarrow& \Big[ \left( a_1 \psi'' + b_1 \psi' + c_1 \psi \right) - \left( a_2 \chi'' + b_2 \chi' + c_2 \chi \right) \Big] + i \Big[ \left( a_2 \psi'' + b_2 \psi' + c_2 \psi \right) + \left( a_1 \chi'' + b_1 \chi' + c_1 \chi \right) \Big] &= 0 \end{align} $$ so it gives a coupled second-order differential equations.

 

Question. How about a first-order differential equation with complex constant coefficients? $$ \begin{align} (a_1 + ia_2) \frac{ \textrm{d} y }{ \textrm{d}x } + (b_1 + ib_2) y &= 0 \end{align} $$

Method 1. Using matrix notation: Let $$ \begin{align} y(x) = \psi(x) + i \chi(x) \end{align} $$ Substituting this into the differential equation gives $$ \begin{align} (a_1 + ia_2) \left( \psi' + i \chi' \right) + (b_1 + ib_2) \left( \psi + i \chi \right) &= 0 \\ \Big[ \left( a_1 \psi' + b_1 \psi \right) - \left( a_2 \chi' + b_2 \chi \right) \Big] + i \Big[ \left( a_2 \psi' + b_2 \psi \right) + \left( a_1 \chi' + b_1 \chi \right) \Big] &= 0 \end{align} $$ The real and imaginary parts yield a set of simultaneous equations $$ \begin{align} \left\{ \begin{array}{l} a_1 \psi' + b_1 \psi = a_2 \chi' + b_2 \chi \\ a_2 \psi' + b_2 \psi = - a_1 \chi' - b_1 \chi \end{array} \right. \qquad &\textrm{or} \qquad \left\{ \begin{array}{l} a_1 \psi' - a_2 \chi' = - b_1 \psi + b_2 \chi \\ a_2 \psi' + a_1 \chi' = - b_2 \psi - b_1 \chi \end{array} \right. \\ \\ \begin{pmatrix} a_1 & b_1 \\ a_2 & b_2 \end{pmatrix} \begin{pmatrix} \psi' \\ \psi \end{pmatrix} = \begin{pmatrix} a_2 & b_2 \\ - a_1 & - b_1 \end{pmatrix} \begin{pmatrix} \chi' \\ \chi \end{pmatrix} \qquad &\textrm{or} \qquad \begin{pmatrix} a_1 & - a_2 \\ a_2 & a_1 \end{pmatrix} \frac{ \textrm{d} }{ \textrm{d}x } \begin{pmatrix} \psi \\ \chi \end{pmatrix} = \begin{pmatrix} - b_1 & b_2 \\ - b_2 & - b_1 \end{pmatrix} \begin{pmatrix} \psi \\ \chi \end{pmatrix} \end{align} $$

The latter appears to be more useful. $$ \begin{align} \begin{pmatrix} a_1 & - a_2 \\ a_2 & a_1 \end{pmatrix} \frac{ \textrm{d} }{ \textrm{d}x } \begin{pmatrix} \psi \\ \chi \end{pmatrix} &= - \begin{pmatrix} b_1 & - b_2 \\ b_2 & b_1 \end{pmatrix} \begin{pmatrix} \psi \\ \chi \end{pmatrix} \\ \\ \Rightarrow \qquad \frac{ \textrm{d} }{ \textrm{d}x } \begin{pmatrix} \psi \\ \chi \end{pmatrix} &= - \begin{pmatrix} a_1 & - a_2 \\ a_2 & a_1 \end{pmatrix}^{-1} \begin{pmatrix} b_1 & - b_2 \\ b_2 & b_1 \end{pmatrix} \begin{pmatrix} \psi \\ \chi \end{pmatrix} \\ &= - \frac1{ a_1^2 + a_2^2 } \begin{pmatrix} a_1 & a_2 \\ - a_2 & a_1 \end{pmatrix} \begin{pmatrix} b_1 & - b_2 \\ b_2 & b_1 \end{pmatrix} \begin{pmatrix} \psi \\ \chi \end{pmatrix} \\ &= - \frac1{ a_1^2 + a_2^2 } \begin{pmatrix} a_1 b_1 + a_2 b_2 & - ( a_1 b_2 - a_2 b_1 ) \\ a_1 b_2 - a_2 b_1 & a_1 b_1 + a_2 b_2 \end{pmatrix} \begin{pmatrix} \psi \\ \chi \end{pmatrix} \end{align} $$ Using the vector notation, $$ \begin{align} \textbf a = \begin{pmatrix} a_1 \\ a_2 \end{pmatrix} \qquad \textrm{and} \qquad \textbf b = \begin{pmatrix} b_1 \\ b_2 \end{pmatrix} \end{align} $$ we can express $$ \begin{align} \textbf a \cdot \textbf b &= a_1 b_1 + a_2 b_2 = p \\ (\textbf a \times \textbf b) \cdot \textbf k &= a_1 b_2 - a_2 b_1 = q \end{align} $$ Then the coupled differential equation reads $$ \begin{align} && \frac{ \textrm{d} }{ \textrm{d}x } \begin{pmatrix} \psi \\ \chi \end{pmatrix} &= - \frac1{ | \textbf a |^2 } \begin{pmatrix} \textbf a \cdot \textbf b & - (\textbf a \times \textbf b) \cdot \textbf k \\ (\textbf a \times \textbf b) \cdot \textbf k & \textbf a \cdot \textbf b \end{pmatrix} \begin{pmatrix} \psi \\ \chi \end{pmatrix} \\ &&&= - \frac1{ | \textbf a |^2 } \begin{pmatrix} p & - q \\ q & p \end{pmatrix} \begin{pmatrix} \psi \\ \chi \end{pmatrix} \\ &&&= - \frac1{ | \textbf a |^2 } \underbrace{ \begin{pmatrix} -\frac{i}{\sqrt{2}} & \frac{i}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix} }_{ P^{-1} } \underbrace{ \begin{pmatrix} p - iq & 0 \\ 0 & p + iq \end{pmatrix} }_{D} \underbrace{ \begin{pmatrix} \frac{i}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ -\frac{i}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix} }_{ P } \begin{pmatrix} \psi \\ \chi \end{pmatrix} \\ &&&= - \frac1{ | \textbf a |^2 } P^{-1}DP \begin{pmatrix} \psi \\ \chi \end{pmatrix} \\ \\ &\Rightarrow& P \begin{pmatrix} \psi' \\ \chi' \end{pmatrix} &= - \frac1{ | \textbf a |^2 } D P \begin{pmatrix} \psi \\ \chi \end{pmatrix} \\ &\Rightarrow& \begin{pmatrix} \frac{i}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ -\frac{i}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix} \begin{pmatrix} \psi' \\ \chi' \end{pmatrix} &= - \frac1{ | \textbf a |^2 } \begin{pmatrix} p - iq & 0 \\ 0 & p + iq \end{pmatrix} \begin{pmatrix} \frac{i}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ -\frac{i}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix} \begin{pmatrix} \psi \\ \chi \end{pmatrix} \\ \end{align} $$ Let $$ \begin{align} \Psi = \frac{ -i \psi + \chi }{ \sqrt{2} } \quad \left[ = -\frac{i}{ \sqrt{2} } (\psi + i \chi) = -\frac{i}{ \sqrt{2} }y \right] \end{align} $$ then the differential equation can be written as $$ \begin{align} && \frac{ \textrm{d} }{ \textrm{d}x } \begin{pmatrix} \Psi^* \\ \Psi \end{pmatrix} &=- \frac1{ | \textbf a |^2 } \begin{pmatrix} p - iq & 0 \\ 0 & p + iq \end{pmatrix} \begin{pmatrix} \Psi^* \\ \Psi \end{pmatrix} \\ \\ &\Rightarrow& \frac{ \textrm d \Psi }{ \textrm dx } &= - \frac{ p + i q }{ | \textbf a |^2 } \Psi \\ &\Rightarrow& \Psi &= A \textrm e^{ - \frac{ p + i q }{ | \textbf a |^2 } x } \\ &&&= A \textrm e^{ - \frac{ \textbf a \cdot \textbf b + i (\textbf a \times \textbf b) \cdot \textbf k }{ | \textbf a |^2 } x } \\ &&&= A \textrm e^{ - \frac{ \textbf a \cdot \textbf b }{ | \textbf a |^2 } x } \left[ \cos \left( \frac{ ( \textbf a \times \textbf b ) \cdot \textbf k }{ | \textbf a |^2 } x \right) - i \sin \left( \frac{ ( \textbf a \times \textbf b ) \cdot \textbf k }{ | \textbf a |^2 } x \right) \right] \end{align} $$ Finally, $$ \begin{align} y &\equiv \psi + i \chi \\ &= i \sqrt{2} \Psi \\ &= i \sqrt{2} A \textrm e^{ - \frac{ \textbf a \cdot \textbf b }{ | \textbf a |^2 } x } \left[ \cos \left( \frac{ ( \textbf a \times \textbf b ) \cdot \textbf k }{ | \textbf a |^2 } x \right) - i \sin \left( \frac{ ( \textbf a \times \textbf b ) \cdot \textbf k }{ | \textbf a |^2 } x \right) \right] \end{align} $$

 

In addition, if we let $$ \begin{align} A = \frac{ \lambda - i \mu }{ \sqrt{2} } \in \mathbb C \end{align} $$ and it gives $$ \begin{align} &\Rightarrow& y &= ( i \lambda + \mu ) \textrm e^{ - \frac{ \textbf a \cdot \textbf b }{ | \textbf a |^2 } x } \left[ \cos \left( \frac{ ( \textbf a \times \textbf b ) \cdot \textbf k }{ | \textbf a | } x \right) - i \sin \left( \frac{ ( \textbf a \times \textbf b ) \cdot \textbf k }{ | \textbf a |^2 } x \right) \right] \\ \\ &\Rightarrow& \psi &= \textrm{Re}(y) = \textrm e^{ - \frac{ \textbf a \cdot \textbf b }{ | \textbf a |^2 } x } \left[ \mu \cos \left( \frac{ ( \textbf a \times \textbf b ) \cdot \textbf k }{ | \textbf a |^2 } x \right) + \lambda \sin \left( \frac{ ( \textbf a \times \textbf b ) \cdot \textbf k }{ | \textbf a |^2 } x \right) \right] \\ && \chi &= \textrm{Im}(y) = \textrm e^{ - \frac{ \textbf a \cdot \textbf b }{ | \textbf a |^2 } x } \left[ \lambda \cos \left( \frac{ ( \textbf a \times \textbf b ) \cdot \textbf k }{ | \textbf a |^2 } x \right) - \mu \sin \left( \frac{ ( \textbf a \times \textbf b ) \cdot \textbf k }{ | \textbf a |^2 } x \right) \right] \end{align}$$

 

Method 2. As an ordinary DE with complex-valued function: For $y \in \mathbb C$, $$ \begin{align} a \frac{ \textrm{d} y }{ \textrm{d}x } + b y &= 0 \qquad \Rightarrow \qquad y = C \textrm e^{ - \frac{b}{a} x } \end{align} $$ where $$ \begin{align} a&= a_1 + i a_2 \\ b&= b_1 + i b_2 \end{align} $$ For the exponent, we find: $$ \begin{align} \frac{b}{a} &= \frac{ a^*b }{ a^*a } \\ &= \frac{ ( a_1 - i a_2 ) ( b_1 + i b_2 ) }{ a_1^2 + a_2^2 } \\ &= \frac{ ( a_1 b_1 + a_2 b_2 ) + i ( a_1 b_2 - a_2 b_1 ) }{ a_1^2 + a_2^2 } \\ &= \frac{ ( \textbf a \cdot \textbf b ) + i ( \textbf a \times \textbf b ) \cdot \textbf k }{ | \textbf a |^2 } \end{align} $$ which gives $$ \begin{align} \Rightarrow \qquad y &= C \textrm e^{ - \frac{ ( \textbf a \cdot \textbf b ) + i ( \textbf a \times \textbf b ) \cdot \textbf k }{ | \textbf a |^2 } x } \\ &= C \textrm e^{ - \frac{ \textbf a \cdot \textbf b }{ | \textbf a |^2 } x } \left[ \cos \left( \frac{ ( \textbf a \times \textbf b ) \cdot \textbf k }{ | \textbf a |^2 } x \right) - i \sin \left( \frac{ ( \textbf a \times \textbf b ) \cdot \textbf k }{ | \textbf a |^2 } x \right) \right] \end{align} $$ If we identify $C = i \sqrt{2} A \in \mathbb C$, it agrees with the result obtained above.