CP2 §7.4 Using boundary conditions

Core pure mathematics 2

Table of contents

1. Introduction
2. Examples
3. Exercise 7D

1. Introduction

We can use given boundary conditions to find a particular solution to a second-order differential equation. Since there are two arbitrary constants, we will need two boundary conditions to determine the complete particular solution.

2. Examples

Example 1. Find $y$ in terms of $x$, given that $$ \begin{align} \frac{ \textrm{d}^2 y }{ \textrm dx^2 } - y = 2 \textrm e^x \end{align} $$ and that $ \frac{ \textrm{d} y }{ \textrm d x } = 0 $ and $ y = 0 $ at $ x = 0 $.

 

 

Solution.

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

 

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

 

(iii) Particular solution: We apply the boundary conditions, $$ \begin{align} && y(x=0) &= A + B = 0 \\ \\ && \frac{ \textrm{d} y }{ \textrm d x } &= A \textrm e^{ x } - B \textrm e^{ -x } + \textrm e^{ x } + x \textrm e^{ x } \\ &\Rightarrow& \left. \frac{ \textrm{d} y }{ \textrm d x } \right\vert_{x=0} &= A - B + 1 = 0 \\ \end{align} $$ i.e. $$ \begin{align} \left\{ \begin{array}{l} A + B = 0 \\ A - B = -1 \end{array} \right. \qquad \Rightarrow \qquad \left\{ \begin{array}{l} A = - \frac12 \\ B = \frac12 \end{array} \right. \end{align} $$ Thus, the particular solution reads $$ \begin{align} y = -\frac12 \textrm e^{ x } + \frac12 \textrm e^{ -x } + x \textrm e^{ x } \qquad \checkmark \end{align} $$

 

Example 2. Given that a particular integral is of the form $ \lambda \sin 2t $, find the solution to the differential equation $$ \begin{align} \frac{ \textrm{d}^2 x }{ \textrm d t^2 } + x = 3 \sin 2t \end{align} $$ for which $ x = 0 $ and $ \frac{ \textrm{d} x }{ \textrm d t } = 1 $ when $ t = 0 $.

 

Solution.

(i) Complementary function: Let $ x = 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 $$ \begin{align} x_1 = A \cos t + B \sin t \end{align} $$

 

(ii) Particular integral: We try the given function, $ \lambda \sin 2t $, which is part of the standard form - see the comment below. $$ \begin{align} x_2 &= \lambda \sin 2t \\ x_2' &= 2 \lambda \cos 2t \\ x_2'' &= -4 \lambda \sin 2t \end{align} $$ Substituting these into the differential equation gives $$ \begin{align} x_2'' + x_2 &= -4 \lambda \sin 2t + \lambda \sin 2t \\ &= -3 \lambda \sin 2t \\ &\stackrel{!}{=} 3 \lambda \sin 2t \end{align} $$ We find: $$ \begin{align} \lambda = -1 \end{align} $$ and the particular integral is $$ \begin{align} x_2 &= - \sin 2t \end{align} $$ Finally, the general solution reads $$ \begin{align} x(t) = x_1 + x_2 = A \cos t + B \sin t - \sin 2t \end{align} $$

Comment. The standard form for the particular integral is $ \lambda \sin 2t + \mu \cos 2t $. $$ \begin{align} x_2 &= \lambda \sin 2t + \mu \cos 2t \\ x_2' &= 2 \lambda \cos 2t - 2 \mu \sin 2t \\ x_2'' &= -4 \lambda \sin 2t - 4 \mu \cos 2t \end{align} $$ Substituting these into the differential equation gives $$ \begin{align} x_2'' + x_2 &= ( -4 \lambda \sin 2t - 4 \mu \cos 2t ) + ( \lambda \sin 2t + \mu \cos 2t ) \\ &= -3 \lambda \sin 2t - 3 \mu \cos 2t \\ &\stackrel{!}{=} 3 \lambda \sin 2t \end{align} $$ so we find that $\mu = 0$, which motivates the particular integral given in the question.

 

(iii) Particular solution: We apply the boundary conditions, $$ \begin{align} && x(0) &= A = 0 \\ \\ && x'(t) &= \underbrace{ - A \sin t }_{ = 0 } + B \cos t - 2 \cos 2t \\ &\Rightarrow& x'(0) &= B - 2 = 1 \qquad \Rightarrow \qquad B = 3 \end{align} $$ Thus, the particular solution reads $$ \begin{align} x(t) = 3 \sin t - \sin 2t \qquad \checkmark \end{align} $$

 

3. Exercise 7D

Question 1. (E)

(a) Find the general solution to the differential equation [5 marks] $$ \begin{align} \frac{ \textrm{d}^2 y }{ \textrm d x^2 } + 5 \frac{ \textrm{d} y }{ \textrm d x } + 6 y = 12 \textrm e^x \end{align} $$

 

(b) Hence find the particular solution that satisfies $ y = 1 $ and $ \frac{ \textrm{d} y }{ \textrm d x } = 0 $ when $ x = 0 $. [4 marks]

 

Solution.

(a) (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: We try the standard form, i.e. $$ \begin{align} y_2 &= \lambda \textrm e^{ x } \\ y_2' &= \lambda \textrm e^{ x } \\ y_2'' &= \lambda \textrm e^{ x } \end{align} $$ Substituting these into the differential equation gives $$ \begin{align} y_2'' + 5 y_2' + 6 y_2 &= \lambda \textrm e^{ x } + 5 \lambda \textrm e^{ x } + 6 \lambda \textrm e^{ x } \\ &= 12 \lambda \textrm e^{ x } \\ &\stackrel{!}{=} 12 \textrm e^{ x } \end{align} $$ We find: $$ \begin{align} \lambda = 1 \end{align} $$ and the particular integral is $$ \begin{align} y_2 &= \textrm e^{ x } \end{align} $$ Finally, the general solution reads $$ \begin{align} y = y_1 + y_2 = A \textrm e^{ -2x } + B \textrm e^{ -3x } + \textrm e^{ x } \qquad \checkmark \end{align} $$

 

(b) Particular solution: We apply the boundary conditions, $$ \begin{align} && y(x=0) &= A + B + 1 = 1 \\ \\ && \frac{ \textrm{d} y }{ \textrm d x } &= -2A \textrm e^{ -2x } - 3B \textrm e^{ -3x } + \textrm e^{ x } \\ &\Rightarrow& \left. \frac{ \textrm{d} y }{ \textrm d x } \right\vert_{x=0} &= -2A - 3B + 1 = 0 \end{align} $$ i.e. $$ \begin{align} \left\{ \begin{array}{l} A + B = 0 \\ 2A + 3B = 1 \end{array} \right. \qquad \Rightarrow \qquad \left\{ \begin{array}{l} A = -1 \\ B = 1 \end{array} \right. \end{align} $$ Thus, the particular solution reads $$ \begin{align} y = - \textrm e^{ -2x } + \textrm e^{ -3x } + \textrm e^{ x } \qquad \checkmark \end{align} $$

 

Question 2. (E)

(a) Find the general solution to the differential equation [5 marks] $$ \begin{align} \frac{ \textrm{d}^2 y }{ \textrm d x^2 } + 2 \frac{ \textrm{d} y }{ \textrm d x } = 12 \textrm e^{ 2x } \end{align} $$

 

(b) Hence find the particular solution that satisfies $ y = 2 $ and $ \frac{ \textrm{d} y }{ \textrm d x } = 6 $ when $ x = 0 $. [5 marks]

 

Solution.

(a) (i) Complementary function: Let $ y = A \textrm e^{ mx }$ and the auxiliary equation reads $$ \begin{align} && m^2 + 2 m &= 0 \\ &\Rightarrow& m ( m + 2 ) &= 0 \\ &\Rightarrow& m = 0 \quad \textrm{or} \quad m&= -2 \end{align} $$ 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: We try the standard form, i.e. $$ \begin{align} y_2 &= \lambda \textrm e^{ 2x } \\ y_2' &= 2 \lambda \textrm e^{ 2x } \\ y_2'' &= 4 \lambda \textrm e^{ 2x } \end{align} $$ Substituting these into the differential equation gives $$ \begin{align} y_2'' + 2 y_2' &= 4 \lambda \textrm e^{ 2x } + 2 \left( 2 \lambda \textrm e^{ 2x } \right) \\ &= 8 \lambda \textrm e^{ 2x } \\ &\stackrel{!}{=} 12 \textrm e^{ 2x } \end{align} $$ We find: $$ \begin{align} \lambda = \frac32 \end{align} $$ and the particular integral is $$ \begin{align} y_2 &= \frac32 \textrm e^{ 2x } \end{align} $$ Finally, the general solution reads $$ \begin{align} y = y_1 + y_2 = A + B \textrm e^{ -2x } + \frac32 \textrm e^{ 2x } \qquad \checkmark \end{align} $$

 

(b) Particular solution: We apply the boundary conditions, $$ \begin{align} && y(x=0) &= A + B + \frac32 = 2 \\ \\ && \frac{ \textrm{d} y }{ \textrm d x } &= - 2B \textrm e^{ -2x } + 3 \textrm e^{ 2x } \\ &\Rightarrow& \left. \frac{ \textrm{d} y }{ \textrm d x } \right\vert_{x=0} &= -2B + 3 = 6 \end{align} $$ i.e. $$ \begin{align} B = -\frac32 \qquad \textrm{and} \qquad A = \frac32 - B = 2 \end{align} $$ Thus, the particular solution reads $$ \begin{align} y = 2 - \frac32 \textrm e^{ -2x } + \frac32 \textrm e^{ 2x } \qquad \checkmark \end{align} $$

 

Question 3. (E) Given that $ y = 0 $ and $ \frac{ \textrm{d} y }{ \textrm d x } = \frac16 $ when $ x = 0 $, find the particular solution to the differential equation

$$ \begin{align} \frac{ \textrm{d}^2 y }{ \textrm d x^2 } - \frac{ \textrm{d} y }{ \textrm d x } - 42 y = 14 \end{align} $$ [10 marks]

 

Solution.

(i) Complementary function: Let $ y = A \textrm e^{ mx }$ and the auxiliary equation reads $$ \begin{align} && m^2 - m - 42 &= 0 \\ &\Rightarrow& ( m - 7 )( m + 6 )&= 0 \\ &\Rightarrow& m = 7 \quad \textrm{or} \quad m&= -6 \end{align} $$ and the complementary function is $$ \begin{align} y_1 = A \textrm e^{ 7x } + B \textrm e^{ -6x } \end{align} $$

 

(ii) Particular integral: We try the standard form, i.e. $$ \begin{align} y_2 &= \lambda \\ y_2' &= 0 \\ y_2'' &= 0 \end{align} $$ Substituting these into the differential equation gives $$ \begin{align} y_2'' - y_2' - 42 y_2 &= -42 \lambda &\stackrel{!}{=} 14 \end{align} $$ and the particular integral is $$ \begin{align} y_2 &= \lambda = -\frac13 \end{align} $$ Finally, the general solution reads $$ \begin{align} y = y_1 + y_2 = A \textrm e^{ 7x } + B \textrm e^{ -6x } - \frac13 \qquad \checkmark \end{align} $$

 

(iii) Particular solution: We apply the boundary conditions, $$ \begin{align} && y(x=0) &= A + B - \frac13 = 0 \\ \\ && \frac{ \textrm{d} y }{ \textrm d x } &= 7A \textrm e^{ 7x } - 6B \textrm e^{ -6x } \\ &\Rightarrow& \left. \frac{ \textrm{d} y }{ \textrm d x } \right\vert_{x=0} &= 7A - 6B = \frac16 \end{align} $$ i.e. $$ \begin{align} \left\{ \begin{array}{l} A + B = \frac13 \\ 7A - 6B = \frac16 \end{array} \right. \qquad \Rightarrow \qquad \left\{ \begin{array}{l} A = \frac1{13} \left( \frac16 + 2 \right) = \frac{1}{6} \\ B = \frac13 - A = \frac13 - \frac{1}{6} = \frac{1}{6} \end{array} \right. \end{align} $$ Thus, the particular solution reads $$ \begin{align} y = \frac{1}{6} \textrm e^{ 7x } + \frac{1}{6} \textrm e^{ -6x } - \frac13 \qquad \checkmark \end{align} $$

 

Question 4. (E)

(a) Find the general solution to the differential equation [6 marks] $$ \begin{align} \frac{ \textrm{d}^2 y }{ \textrm d x^2 } + 9 y = 16 \sin x \end{align} $$

 

(b) Hence find the particular solution that satisfies $ y = 1 $ and $ \frac{ \textrm{d} y }{ \textrm d x } = 8 $ when $ x = 0 $. [6 marks]

 

Solution.

(a) (i) Complementary function: Let $ y = A \textrm e^{ mx }$ and the auxiliary equation reads $$ \begin{align} && m^2 + 9 &= 0 \\ &\Rightarrow& m&= \pm 3i \end{align} $$ and the complementary function is $$ \begin{align} y_1 = A \cos 3x + B \sin 3x \end{align} $$

 

(ii) Particular integral: We try the standard form, i.e. $$ \begin{align} y_2 &= \lambda \cos x + \mu \sin x \\ y_2' &= - \lambda \sin x + \mu \cos x \\ y_2'' &= - \lambda \cos x - \mu \sin x \end{align} $$ Substituting these into the differential equation gives $$ \begin{align} y_2'' + 9 y_2 &= ( - \lambda \cos x - \mu \sin x ) + 9 ( \lambda \cos x + \mu \sin x ) \\ &= 8 \lambda \cos x + 8 \mu \sin x \\ &\stackrel{!}{=} 16 \sin x \end{align} $$ We find: $$ \begin{align} \lambda = 0 \qquad \textrm{and} \qquad \mu = 2 \end{align} $$ and the particular integral is $$ \begin{align} y_2 &= 2 \sin x \end{align} $$ Finally, the general solution reads $$ \begin{align} y = y_1 + y_2 = A \cos 3x + B \sin 3x + 2 \sin x \qquad \checkmark \end{align} $$

 

(b) Particular solution: We apply the boundary conditions, $$ \begin{align} && y(x=0) &= A = 1 \\ \\ && \frac{ \textrm{d} y }{ \textrm d x } &= -3A \sin 3x + 3B \cos 3x + 2 \cos x \\ &\Rightarrow& \left. \frac{ \textrm{d} y }{ \textrm d x } \right\vert_{x=0} &= 3B + 2 = 8 \qquad \Rightarrow \qquad B = 2 \end{align} $$ Thus, the particular solution reads $$ \begin{align} y = \cos 3x + 2 \sin 3x + 2 \sin x \qquad \checkmark \end{align} $$

 

Question 5. (E)

(a) Find the general solution to the differential equation [6 marks] $$ \begin{align} 4 \frac{ \textrm{d}^2 y }{ \textrm d x^2 } + 4 \frac{ \textrm{d} y }{ \textrm d x } + 5 y = \sin x + 4 \cos x \end{align} $$

 

(b) Hence find the particular solution that satisfies $ y = 0 $ and $ \frac{ \textrm{d} y }{ \textrm d x } = 0 $ when $ x = 0 $. [6 marks]

 

Solution.

(a) (i) Complementary function: Let $ y = A \textrm e^{ mx }$ and the auxiliary equation reads $$ \begin{align} && 4m^2 + 4m + 5 &= 0 \\ && 4 \left( m^2 + m \right) + 5 &= 0 \\ &\Rightarrow& 4 \left[ \left( m + \frac12 \right)^2 - \frac14 \right] + 5 &= 0 \\ &\Rightarrow& 4 \left( m + \frac12 \right)^2 + 4 &= 0 \\ &\Rightarrow& m&= -\frac12 \pm i \end{align} $$ and the complementary function is $$ \begin{align} y_1 = \textrm e^{ - \frac12 x } ( A \cos x + B \sin x ) \end{align} $$

 

(ii) Particular integral: We try the standard form, i.e. $$ \begin{align} y_2 &= \lambda \cos x + \mu \sin x \\ y_2' &= - \lambda \sin x + \mu \cos x \\ y_2'' &= - \lambda \cos x - \mu \sin x \end{align} $$ Substituting these into the differential equation gives $$ \begin{align} 4 y_2'' + 4 y_2' + 5 y_2 &= 4 ( - \lambda \cos x - \mu \sin x ) + 4 ( - \lambda \sin x + \mu \cos x ) + 5 ( \lambda \cos x + \mu \sin x ) \\ &= ( \lambda + 4 \mu ) \cos x + ( - 4 \lambda + \mu ) \sin x \\ &\stackrel{!}{=} \sin x + 4 \cos x \end{align} $$ i.e. $$ \begin{align} \left\{ \begin{array}{l} \lambda + 4 \mu = 4 \\ - 4 \lambda + \mu = 1 \end{array} \right. \qquad \Rightarrow \qquad \left\{ \begin{array}{l} \lambda = 0 \\ \mu = 1 \end{array} \right. \end{align} $$ and the particular integral is $$ \begin{align} y_2 &= \sin x \end{align} $$ Finally, the general solution reads $$ \begin{align} y = y_1 + y_2 = \textrm e^{ - \frac12 x } ( A \cos x + B \sin x ) + \sin x \qquad \checkmark \end{align} $$

 

(b) Particular solution: We apply the boundary conditions, $$ \begin{align} && y(x=0) &= A = 0 \\ \\ && \frac{ \textrm{d} y }{ \textrm d x } &= - \frac12 B \textrm e^{ - \frac12 x } \sin x + B \textrm e^{ - \frac12 x } \cos x + \cos x \\ &\Rightarrow& \left. \frac{ \textrm{d} y }{ \textrm d x } \right\vert_{x=0} &= B + 1 = 0 \qquad \Rightarrow \qquad B = -1 \end{align} $$ Thus, the particular solution reads $$ \begin{align} y = - \textrm e^{ - \frac12 x } \sin x + \sin x = \left( 1 - \textrm e^{ - \frac12 x } \right) \sin x \qquad \checkmark \end{align} $$

 

Question 6. (E/P)

(a) Find the general solution to the differential equation [6 marks] $$ \begin{align} \frac{ \textrm{d}^2 x }{ \textrm d t^2 } - 3 \frac{ \textrm{d} x }{ \textrm d t } + 2 x = 2t - 3 \end{align} $$

 

(b) Given that $ x = 1 $ when $ t = 0 $, and $ x = 2 $ when $ t = 1 $, find a particular solution of this differential equation. [6 marks]

 

Solution.

(a) (i) Complementary function: Let $ x = A \textrm e^{ mt }$ and the auxiliary equation reads $$ \begin{align} && m^2 - 3m + 2 &= 0 \\ && ( m - 1 )( m - 2) &= 0 \\ &\Rightarrow& m = 1 \quad \textrm{or} \quad m&= 2 \end{align} $$ and the complementary function is $$ \begin{align} x_1 = A \textrm e^{ t } + B \textrm e^{ 2t } \end{align} $$

 

(ii) Particular integral: We try the standard form, i.e. $$ \begin{align} x_2 &= \lambda + \mu t \\ x_2' &= \mu \\ x_2'' &= 0 \end{align} $$ Substituting these into the differential equation gives $$ \begin{align} x_2'' - 3 x_2' + 2 x_2 &= - 3 \mu + 2 ( \lambda + \mu t ) \\ &= ( 2 \lambda - 3 \mu ) + 2 \mu t \\ &\stackrel{!}{=} 2t - 3 \end{align} $$ i.e. $$ \begin{align} \left\{ \begin{array}{l} 2 \mu = 2 \\ 2 \lambda - 3 \mu = -3 \end{array} \right. \qquad \Rightarrow \qquad \left\{ \begin{array}{l} \mu = 1 \\ \lambda = \frac{ 3 \mu - 3 }{ 2 } = 0 \end{array} \right. \end{align} $$ and the particular integral is $$ \begin{align} x_2 &= t \end{align} $$ Finally, the general solution reads $$ \begin{align} x(t) = x_1 + x_2 = A \textrm e^{ t } + B \textrm e^{ 2t } + t \qquad \checkmark \end{align} $$

 

(b) Particular solution: We apply the boundary conditions, $$ \begin{align} x(t=0) &= A + B = 1 \\ \\ x(t=1) &= A \textrm e + B \textrm e^2 + 1 = 2 \end{align} $$ which gives $$ \begin{align} \left\{ \begin{array}{l} A + B = 1 \\ A \textrm e + B \textrm e^2 = 1 \end{array} \right. \qquad \Rightarrow \qquad \left\{ \begin{array}{l} B = \frac{ 1 - \textrm e }{ \textrm e^2 - \textrm e } = \frac{ 1 - \textrm e }{ \textrm e ( \textrm e - 1 ) } = - \frac1{ \textrm e } \\ A = 1 - B = 1 + \frac1{ \textrm e } = \frac{ 1 + \textrm e }{ \textrm e } \end{array} \right. \end{align} $$ Thus, the particular solution reads $$ \begin{align} x(t) &= \left( 1 + \frac1{ \textrm e } \right) \textrm e^{ t } - \frac1{ \textrm e } \textrm e^{ 2t } + t \\ &= \left( 1 + \textrm e \right) \textrm e^{ t - 1 } - \frac1{ \textrm e } \textrm e^{ 2t } + t \\ &= \textrm e^{ t } + \textrm e^{ t - 1 } - \textrm e^{ 2t - 1 } + t \qquad \checkmark \end{align} $$

 

Question 7. (E) Find the particular solution to the differential equation $$ \begin{align} \frac{ \textrm{d}^2 x }{ \textrm d t^2 } - 9 x = 10 \sin t \end{align} $$ that satisfies $ x = 2 $ and $ \frac{ \textrm{d} x }{ \textrm d t } = - 1 $ when $ t = 0 $. [10 marks]

 

Solution.

(i) Complementary function: Let $ x = A \textrm e^{ mt }$ and the auxiliary equation reads $$ \begin{align} && m^2 - 9 &= 0 \\ &\Rightarrow& m&= \pm 3 \end{align} $$ and the complementary function is $$ \begin{align} x_1 = A \textrm e^{ 3t } + B \textrm e^{ -3t } \end{align} $$

 

(ii) Particular integral: We try the standard form.

 

Comment: Due to the absence of the $\frac{ \textrm{d} x }{ \textrm d t }$-term in the equation, the particular integral would be of the form $ \mu \sin t$. However, we still try the standard form including the cosine term to illustrate a general approach. See also Question 12 below.

 

$$ \begin{align} x_2 &= \lambda \cos t + \mu \sin t \\ x_2' &= - \lambda \sin t + \mu \cos t \\ x_2'' &= - \lambda \cos t - \mu \sin t \end{align} $$ Substituting these into the differential equation gives $$ \begin{align} x_2'' - 9 x_2 &= ( - \lambda \cos t - \mu \sin t ) - 9 ( \lambda \cos t + \mu \sin t ) \\ &= - 10 \lambda \cos t - 10 \mu \sin t \\ &\stackrel{!}{=} 10 \sin t \end{align} $$ i.e. $$ \begin{align} \lambda = 0 \qquad \textrm{and} \qquad \mu = -1 \end{align} $$ and the particular integral is $$ \begin{align} x_2 &= - \sin t \end{align} $$ Finally, the general solution reads $$ \begin{align} x(t) = x_1 + x_2 = A \textrm e^{ 3t } + B \textrm e^{ -3t } - \sin t \qquad \checkmark \end{align} $$

 

(iii) Particular solution: We apply the boundary conditions, $$ \begin{align} && x(t=0) &= A + B = 2 \\ \\ && \frac{ \textrm{d} x }{ \textrm d t } &= 3 A \textrm e^{ 3t } - 3 B \textrm e^{ -3t } - \cos t \\ &\Rightarrow& \left. \frac{ \textrm{d} x }{ \textrm d t } \right\vert_{t=0} &= 3 A - 3 B - 1 = - 1 \end{align} $$ which gives $$ \begin{align} \left\{ \begin{array}{l} A + B = 2 \\ A - B = 0 \end{array} \right. \qquad \Rightarrow \qquad A = B = 1 \end{align} $$ Thus, the particular solution reads $$ \begin{align} x(t) &= \textrm e^{ 3t } + \textrm e^{ -3t } - \sin t = 2 \cosh 3t - \sin t \qquad \checkmark \end{align} $$

 

Question 8. (E/P)

(a) (i) Find the value of $ \lambda $ for which $ x = \lambda t^3 \textrm e^{ 2t } $ is a particular solution to the differential equation $$ \begin{align} \frac{ \textrm{d}^2 x }{ \textrm d t^2 } - 4 \frac{ \textrm{d} x }{ \textrm d t } + 4 x = 3 t \textrm e^{ 2t } \end{align} $$ (ii) Hence find the general solution to the differential equation. [6 marks]

 

(b) Find the particular solution that satisfies $ x = 0 $ and $ \frac{ \textrm{d} x }{ \textrm d t } = 1 $ when $ t = 0 $. [6 marks]

 

Solution.

(a) (i) Particular solution: $ x_2 = \lambda t^3 \textrm e^{ 2t } $ $$ \begin{align} x_2 &= \lambda t^3 \textrm e^{ 2t } \\ x_2' &= 3 \lambda t^2 \textrm e^{ 2t } + 2 \lambda t^3 \textrm e^{ 2t } \\ &= \left( 3 t^2 + 2 t^3 \right) \lambda \textrm e^{ 2t } \\ x_2'' &= \left( 6 t + 6 t^2 \right) \lambda \textrm e^{ 2t } + 2 \left( 3 t^2 + 2 t^3 \right) \lambda \textrm e^{ 2t } \\ &= \left( 6 t + 12 t^2 + 4 t^3 \right) \lambda \textrm e^{ 2t } \end{align} $$ Substituting these into the differential equation gives $$ \begin{align} x_2'' - 4 x_2' + 4 x_2 &= \left( 6 t + 12 t^2 + 4 t^3 \right) \lambda \textrm e^{ 2t } - 4 \left( 3 t^2 + 2 t^3 \right) \lambda \textrm e^{ 2t } + 4 \lambda t^3 \textrm e^{ 2t } \\ &= 6 t \lambda \textrm e^{ 2t } + \underbrace{ \left( 12 t^2 - 12 t^2 \right) }_{ = 0 } \lambda \textrm e^{ 2t } + \underbrace{ \left( 4t^3 - 8 t^3 + 4 t^3 \right) }_{ = 0 } \lambda \textrm e^{ 2t } \\ &\stackrel{!}{=} 3 t \textrm e^{ 2t } \end{align} $$ i.e. $$ \begin{align} \lambda = \frac12 \end{align} $$ and the particular integral is $$ \begin{align} x_2 &= \frac12 t^3 \textrm e^{ 2t } \qquad \checkmark \end{align} $$

 

(ii) Complementary function: Let $ x = A \textrm e^{ mt }$ and the auxiliary equation reads $$ \begin{align} && m^2 - 4m + 4 &= 0 \\ && ( m - 2 )^2 &= 0 \\ &\Rightarrow& m&= 2 \quad \textrm{(a repeated root)} \end{align} $$ and the complementary function is $$ \begin{align} x_1 = ( A + B t ) \textrm e^{ 2t } \end{align} $$ Finally, the general solution reads $$ \begin{align} x(t) = x_1 + x_2 = ( A + B t ) \textrm e^{ 2t } + \frac12 t^3 \textrm e^{ 2t } \qquad \checkmark \end{align} $$

 

(b) Particular solution: We apply the boundary conditions, $$ \begin{align} && x(t=0) &= A = 0 \\ \\ && \frac{ \textrm{d} x }{ \textrm d t } &= B \textrm e^{ 2t } + 2 ( A + B t ) \textrm e^{ 2t } + \frac32 t^2 \textrm e^{ 2t } + t^3 \textrm e^{ 2t } \\ &&&= ( 2A + B + 2B t ) \textrm e^{ 2t } + \left( \frac32 t^2 + t^3 \right) \textrm e^{ 2t } \\ &\Rightarrow& \left. \frac{ \textrm{d} x }{ \textrm d t } \right\vert_{t=0} &= B = 1 \end{align} $$ Thus, the particular solution reads $$ \begin{align} x(t) &= t \textrm e^{ 2t } + \frac12 t^3 \textrm e^{ 2t } = \left( t + \frac12 t^3 \right) \textrm e^{ 2t } \qquad \checkmark \end{align} $$

 

Question 9. (E) Find the particular solution to the differential equation $$ \begin{align} 25 \frac{ \textrm{d}^2 x }{ \textrm d t^2 } + 36 x = 18 \end{align} $$ that satisfies $ x = 1 $ and $ \frac{ \textrm{d} x }{ \textrm d t } = 0.6 $ when $ t = 0 $. [12 marks]

 

Solution.

(i) Complementary function: Let $ x = A \textrm e^{ mt }$ and the auxiliary equation reads $$ \begin{align} && 25 m^2 + 36 &= 0 \\ &\Rightarrow& m^2 &= - \frac{ 36 }{ 25 } \\ &\Rightarrow& m&= \pm \frac{6}{5}i \end{align} $$ and the complementary function is $$ \begin{align} x_1 = A \cos \left( \frac{6}{5}t \right) + B \sin \left( \frac{6}{5}t \right) \end{align} $$

 

(ii) Particular integral: We try the standard form, i.e. $$ \begin{align} x_2 &= \lambda \\ x_2' &= 0 \\ x_2'' &= 0 \end{align} $$ Substituting these into the differential equation gives $$ \begin{align} 25 x_2'' + 36 x_2 &= 36 \lambda &\stackrel{!}{=} 18 \end{align} $$ and the particular integral is $$ \begin{align} x_2 &= \lambda = \frac12 \end{align} $$ Finally, the general solution reads $$ \begin{align} x(t) = x_1 + x_2 = A \cos \left( \frac{6}{5}t \right) + B \sin \left( \frac{6}{5}t \right) + \frac12 \qquad \checkmark \end{align} $$

 

(iii) Particular solution: We apply the boundary conditions, $$ \begin{align} && x(t=0) &= A + \frac12 = 1 \\ \\ && \frac{ \textrm{d} x }{ \textrm d t } &= - \frac65 A \sin \left( \frac{6}{5}t \right) + \frac65 B \cos \left( \frac{6}{5}t \right) \\ &\Rightarrow& \left. \frac{ \textrm{d} x }{ \textrm d t } \right\vert_{t=0} &= \frac65 B = \frac35 \end{align} $$ which gives $$ \begin{align} A = B = \frac12 \end{align} $$ Thus, the particular solution reads $$ \begin{align} x(t) &= \frac12 \cos \left( \frac{6}{5}t \right) + \frac12 \sin \left( \frac{6}{5}t \right) + \frac12 \\ &= \frac12 \left( \cos \frac{6}{5}t + \sin \frac{6}{5}t + 1 \right) \qquad \checkmark \end{align} $$

 

Question 10. (E)

(a) Find the general solution to the differential equation $$ \begin{align}   \frac{ \textrm{d}^2 x }{ \textrm d t^2 } - 2 \frac{ \textrm{d} x }{ \textrm d t } + 2 x = 2 t^2   \end{align} $$

 

(b) Hence find the particular solution that satisfies $ x = 1 $ and $ \frac{ \textrm{d} x }{ \textrm d t } = 3 $ when $ t = 0 $. [6 marks]

 

Solution.

(a) (i) Complementary function: Let $ x = A \textrm e^{ mt }$ and the auxiliary equation reads   $$ \begin{align}   && m^2 - 2m + 2 &= 0 \\   &\Rightarrow& ( m - 1 )^2 + 1 &= 0 \\   &\Rightarrow& m&= 1 \pm i   \end{align} $$   and the complementary function is   $$ \begin{align}   x_1 = \textrm e^t ( A \cos t + B \sin t )   \end{align} $$

 

(ii) Particular integral: We try the standard form, i.e.   $$ \begin{align}   x_2 &= \lambda + \mu t + \nu t^2 \\   x_2' &= \mu + 2 \nu t \\   x_2'' &= 2 \nu   \end{align} $$   Substituting these into the differential equation gives   $$ \begin{align}   x_2'' - 2 x_2' + 2 x_2   &= 2 \nu - 2 ( \mu + 2 \nu t ) + 2 ( \lambda + \mu t + \nu t^2 ) \\   &= ( 2 \nu - 2 \mu + 2 \lambda ) + ( -4 \nu + 2 \mu ) t + 2 \nu t^2 \\   &\stackrel{!}{=} 2 t^2   \end{align} $$   i.e.   $$ \begin{align}   \left\{ \begin{array}{l}   2 \nu = 2 \\    -2 \nu + \mu = 0 \\   \nu - \mu + \lambda = 0   \end{array} \right.   \qquad \Rightarrow \qquad   \left\{ \begin{array}{l}   \nu = 1 \\   \mu = 2 \nu = 2 \\   \lambda = \mu - \nu = 2 - 1 = 1   \end{array} \right.   \end{align} $$   and the particular integral is   $$ \begin{align}   x_2 &= 1 + 2 t + t^2   \end{align} $$   Finally, the general solution reads   $$ \begin{align}   x(t) = x_1 + x_2 =  \textrm e^t ( A \cos t + B \sin t ) + 1 + 2 t + t^2   \qquad \checkmark   \end{align} $$

 

(b) Particular solution: We apply the boundary conditions,   $$ \begin{align}   && x(t=0) &= A + 1 = 1 &&\Rightarrow & A &= 0 \\ \\   && \frac{ \textrm{d} x }{ \textrm d t } &=   \textrm e^t ( A \cos t + B \sin t )   + \textrm e^t ( - A \sin t + B \cos t )   + 2 + 2 t \\   &&&= \textrm e^t \Big[ (A+B) \cos t + (B-A) \sin t \Big] + 2 + 2t \\     &\Rightarrow& \left. \frac{ \textrm{d} x }{ \textrm d t } \right\vert_{t=0} &=   A + B + 2 = 3   &&\Rightarrow & B &= 1   \end{align} $$   Thus, the particular solution reads   $$ \begin{align}   x(t) &= \textrm e^t \sin t + 1 + 2 t + t^2 = \textrm e^t \sin t + ( 1 + t )^2   \qquad \checkmark   \end{align} $$

 

Question 11. (E/P)

(a) Find the general solution to the differential equation [7 marks] $$ \begin{align}   \frac{ \textrm{d}^2 y }{ \textrm d x^2 } - 3 \frac{ \textrm{d} y }{ \textrm d x } + 2 y = 3 \textrm e^{ 2x } \end{align} $$

 

(b) Hence find the particular solution that satisfies $ y = 0 $, $ \frac{ \textrm{d} y }{ \textrm d x } = 0 $ when $ x = 0 $. [6 marks]

 

Solution.

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

 

(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   $$ \begin{align}   y_2'' - 3 y_2' + 2 y_2   &= ( 4 + 4x ) \lambda \textrm e^{ 2x } - 3 ( 1 + 2x ) \lambda \textrm e^{ 2x } + 2 \lambda x \textrm e^{ 2x } \\   &= ( 4 - 3 + 4x - 6x + 2x ) \lambda \textrm e^{ 2x } \\   &= \lambda \textrm e^{ 2x } \\   &\stackrel{!}{=} 3 \textrm e^{ 2x }   \end{align} $$   i.e.   $$ \begin{align}     \lambda = 3     \end{align} $$   and the particular integral is   $$ \begin{align}   y_2 &= 3 x \textrm e^{ 2x }   \end{align} $$   Finally, the general solution reads   $$ \begin{align}   y = y_1 + y_2 = A \textrm e^{ x } + B \textrm e^{ 2x } + 3 x \textrm e^{ 2x }   \qquad \checkmark   \end{align} $$

 

(b) Particular solution: We apply the boundary conditions,   $$ \begin{align}   && y(x=0) &= A + B = 0 \\ \\   && \frac{ \textrm{d} y }{ \textrm d x } &=   A \textrm e^{ x } + 2 B \textrm e^{ 2x }   + 3 \textrm e^{ 2x } + 6 x \textrm e^{ 2x } \\     &\Rightarrow& \left. \frac{ \textrm{d} y }{ \textrm d x } \right\vert_{x=0} &=   A + 2B + 3 = 0   \end{align} $$   i.e.   $$ \begin{align}   \left\{ \begin{array}{l}   A + B = 0 \\    A + 2B = -3   \end{array} \right.   \qquad \Rightarrow \qquad   \left\{ \begin{array}{l}   B = -3 \\   A = -B = 3   \end{array} \right.   \end{align} $$   Thus, the particular solution reads   $$ \begin{align}   y &= 3 \textrm e^{ x } - 3 \textrm e^{ 2x } + 3 x \textrm e^{ 2x }   = 3 \left[ \textrm e^{ x } + ( x - 1 ) \textrm e^{ 2x } \right]   \qquad \checkmark   \end{align} $$

 

Question 12. (E/P) Solve the differential equation

$$ \begin{align} \frac{ \textrm{d}^2 y }{ \textrm d x^2 } + 9 y = \sin 3x \end{align} $$ subject to the boundary conditions $ y = 0 $, $ \frac{ \textrm{d} y }{ \textrm d x } = 0 $ when $ x = 0 $. [14 marks]

 

Solution.

(i) Complementary function: Let $ y = A \textrm e^{ mx }$ and the auxiliary equation reads   $$ \begin{align}   && m^2 + 9 &= 0 \\   &\Rightarrow& m&= \pm 3i   \end{align} $$   and the complementary function is   $$ \begin{align}   y_1 = A \cos 3x + B \sin 3x   \end{align} $$

 

(ii) Particular integral: The standard form, $ \lambda \cos 3x + \mu \sin 3x $, is already part of the C.F., so we try the standard form multiplied by $x$.   $$ \begin{align}   y_2 &= x ( \lambda \cos 3x + \mu \sin 3x ) \\   y_2' &= ( \lambda \cos 3x + \mu \sin 3x ) + x ( - 3 \lambda \sin 3x + 3 \mu \cos 3x ) \\   &= ( \lambda + 3 \mu x ) \cos 3x + ( \mu - 3 \lambda x ) \sin 3x \\   y_2'' &= ( - 3 \lambda \sin 3x + 3 \mu \cos 3x ) + ( - 3 \lambda \sin 3x + 3 \mu \cos 3x ) + x ( - 9 \lambda \cos 3x - 9 \mu \sin 3x ) \\   &= ( 6 \mu - 9 \lambda x ) \cos 3x + ( - 6 \lambda - 9 \mu x ) \sin 3x   \end{align} $$   Substituting these into the differential equation gives   $$ \begin{align}   y_2'' + 9 y_2   &= ( 6 \mu - 9 \lambda x ) \cos 3x + ( - 6 \lambda - 9 \mu x ) \sin 3x + 9 x ( \lambda \cos 3x + \mu \sin 3x ) \\   &= ( 6 \mu - 9 \lambda x + 9 \lambda x ) \cos 3x + ( - 6 \lambda - 9 \mu x + 9 \mu x ) \sin 3x \\ &= 6 \mu \cos 3x - 6 \lambda \sin 3x \\   &\stackrel{!}{=} \sin 3x   \end{align} $$   i.e.   $$ \begin{align}     \lambda = -\frac16 \qquad \textrm{and} \qquad \mu = 0     \end{align} $$   and the particular integral is   $$ \begin{align}   y_2 &= - \frac16 x \cos 3x   \end{align} $$ Comment: Due to the absence of the $\frac{ \textrm{d} y }{ \textrm d x }$ term, we may expect to only have $ \mu x \sin 3x $ for the particular integral. However, although we still need only one term for the P.I., it turns out to be the $ \lambda x \cos 3x $ term that we need rather than $ \mu x \sin 3x $. See also Question 7 above.

 

  Finally, the general solution reads   $$ \begin{align}   y = y_1 + y_2 = A \cos 3x + B \sin 3x - \frac16 x \cos 3x   \qquad \checkmark   \end{align} $$

 

(iii) Particular solution: We apply the boundary conditions,   $$ \begin{align}   && y(x=0) &= A = 0 \\ \\   && \frac{ \textrm{d} y }{ \textrm d x } &=   -3 A \sin 3x + 3 B \cos 3x - \frac16 \cos 3x + \frac12 x \sin 3x \\     &\Rightarrow& \left. \frac{ \textrm{d} y }{ \textrm d x } \right\vert_{x=0} &=   3B - \frac16 = 0   \end{align} $$   i.e.   $$ \begin{align}   A = 0 \qquad \textrm{and} \qquad B = \frac1{18}   \end{align} $$   Thus, the particular solution reads   $$ \begin{align}   y &= \frac1{18} \sin 3x - \frac16 x \cos 3x   \qquad \checkmark   \end{align} $$

 

Question 13. (E/P)

$$ \begin{align} \frac{ \textrm{d}^2 x }{ \textrm d t^2 } + 5 \frac{ \textrm{d} x }{ \textrm d t } + 6 x = 2 \textrm e^{ -t } \end{align} $$ Given that $ x = 0 $ and $ \frac{ \textrm{d} x }{ \textrm d t } = 2 $ at $ t = 0 $,

 

(a) find $x$ in terms of $t$. [8 marks]

 

(b) Show that the maximum value of $ x $ is $ \frac{ 2 \sqrt{3} }{ 9 } $ and justify that this is a maximum. [7 marks]

 

Solution.

(a) (i) Complementary function: Let $ x = A \textrm e^{ mt }$ and the auxiliary equation reads   $$ \begin{align}   && m^2 + 5 m + 6 &= 0 \\ &\Rightarrow& ( m + 2 )( m + 3 ) &= 0 \\   &\Rightarrow& m = -2 \quad \textrm{or} \quad m&= -3   \end{align} $$   and the complementary function is   $$ \begin{align}   x_1 = A \textrm e^{ -2t } + B \textrm e^{ -3t }   \end{align} $$

 

(ii) Particular integral: The standard form, $ \lambda \textrm e^{-t} $, i.e.   $$ \begin{align}   x_2 &= \lambda \textrm e^{-t} \\   x_2' &= - \lambda \textrm e^{-t} \\   x_2'' &= \lambda \textrm e^{-t}   \end{align} $$   Substituting these into the differential equation gives   $$ \begin{align}   x_2'' + 5 x_2' + 6 x_2   &= \lambda \textrm e^{-t} - 5 \lambda \textrm e^{-t} + 6 \lambda \textrm e^{-t} \\ &= 2 \lambda \textrm e^{-t} \\   &\stackrel{!}{=} 2 \textrm e^{-t}   \end{align} $$   i.e.   $$ \begin{align}     \lambda = 1     \end{align} $$   and the particular integral is   $$ \begin{align}   x_2 &= \textrm e^{ -t }   \end{align} $$   Finally, the general solution reads   $$ \begin{align}   x(t) = x_1 + x_2 = A \textrm e^{ -2t } + B \textrm e^{ -3t } + \textrm e^{ -t }   \qquad \checkmark   \end{align} $$

 

(iii) Particular solution: We apply the boundary conditions,   $$ \begin{align}   && x(x=0) &= A + B + 1 = 0 \\ \\   && \frac{ \textrm{d} x }{ \textrm d t } &=   -2 A \textrm e^{ -2t } - 3 B \textrm e^{ -3t } - \textrm e^{ -t } \\     &\Rightarrow& \left. \frac{ \textrm{d} x }{ \textrm d t } \right\vert_{t=0} &=   -2 A - 3 B - 1 = 2   \end{align} $$   i.e.   $$ \begin{align}   \left\{ \begin{array}{l}   A + B = -1 \\   2 A + 3 B = -3   \end{array} \right.   \qquad \Rightarrow \qquad   \left\{ \begin{array}{l}   B = -1 \\   A = -1-B = 0   \end{array} \right.   \end{align} $$   Thus, the particular solution reads   $$ \begin{align}   x(t) &= - \textrm e^{ -3t } + \textrm e^{ -t }   \qquad \checkmark   \end{align} $$

 

(b) We find the stationary point(s) by solving $\frac{ \textrm{d} x }{ \textrm d t } = 0$, i.e. $$ \begin{align} && x(t) &= - \textrm e^{ -3t } + \textrm e^{ -t } \\ \\ & \Rightarrow & \frac{ \textrm{d} x }{ \textrm d t } &= 3 \textrm e^{ -3t } - \textrm e^{ -t } \\ &&&= \textrm e^{ -t } \left( 3 \textrm e^{ -2t } - 1 \right) = 0 \end{align} $$ Since $ \textrm e^{ -t } > 0 $ for $ t \in \mathbb R $, we find $$ \begin{align} 3 \textrm e^{ -2t } - 1 = 0 \qquad \Rightarrow \qquad t = - \frac12 \ln \frac13 = \frac12 \ln 3 \end{align} $$ It is convenient to note that $ \textrm e^{-t} = \frac1{\sqrt{3}}$, and it gives $$ \begin{align} x\left( t = \frac12 \ln 3 \right) &= - \left( \frac1{\sqrt{3}} \right)^3 + \left( \frac1{\sqrt{3}} \right) \\ &= - \frac1{ 3 \sqrt{3} } + \frac1{ \sqrt{3} } \\ &= \frac2{ 3 \sqrt{3} } \\ &= \frac{ 2 \sqrt{3} }{ 9 } \qquad \checkmark \end{align} $$ In order to verify the nature of this stationary point, we examine the value of the second-order derivative at $t = \frac12 \ln 3$. $$ \begin{align} && \frac{ \textrm{d} x }{ \textrm d t } &= 3 \textrm e^{ -3t } - \textrm e^{ -t } \\ \\ & \Rightarrow & \frac{ \textrm{d}^2 x }{ \textrm d t^2 } &= - 9 \textrm e^{ -3t } + \textrm e^{ -t } \\ & \Rightarrow & \left. \frac{ \textrm{d}^2 x }{ \textrm d t^2 } \right\vert_{ t = \frac12 \ln 3 } &= - 9 \left( \frac1{\sqrt{3}} \right)^3 + \left( \frac1{\sqrt{3}} \right) \\ &&&= - \frac{ 9 }{ 3 \sqrt{3} } + \frac1{\sqrt{3}} \\ &&&= - \frac{ 2 }{ \sqrt{3} }<0 \end{align} $$ So the stationary point $ (t,x)= \left( \frac12 \ln 3, \frac{ 2 \sqrt{3} }{ 9 } \right) $ is a local maximum.