Q29E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find the solution to the initial value problem.$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{\theta }\right)\mathbf{-}\mathbf{y}\left(\mathbf{\theta }\right)\mathbf{=}\mathbf{sin\theta }\mathbf{-}{\mathbf{e}}^{2\theta }\mathbf{;}\mathbf{y}\left(\mathbf{0}\right)\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\left(\mathbf{0}\right)\mathbf{=}\mathbf{-}\mathbf{1}$

Step-by-Step Solution

Answer

The solution to the initial value problem is $\mathbf{y}\mathbf{=}\frac{3}{4}{\mathbf{e}}^{\theta }\mathbf{+}\frac{7}{12}{\mathbf{e}}^{-\theta }\mathbf{-}\frac{1}{2}\mathbf{sin\theta }\mathbf{-}\frac{1}{3}{\mathbf{e}}^{2\theta }.$

Step 1: Write the auxiliary equation of the given differential equation.

The differential equation is,

 $\mathrm{y}\text{'}\text{'}\left(\mathrm{\theta }\right)-\mathrm{y}\left(\mathrm{\theta }\right)=\mathrm{sin\theta }-{\mathrm{e}}^{2\mathrm{\theta }}......\left(1\right)$

 

Write the homogeneous differential equation of the equation (1),

 $\mathrm{y}\text{'}\text{'}\left(\mathrm{\theta }\right)-\mathrm{y}\left(\mathrm{\theta }\right)=0$

 

The auxiliary equation for the above equation,

 $\begin{array}{c}{\mathrm{m}}^2-1=0\\ \mathrm{m}=\pm 1\end{array}$

Step 2: Find the complementary solution of the given equation.

The root of an auxiliary equation is, 

 

${\mathrm{m}}_1=1,{\mathrm{m}}_2=-1$

 

The complementary solution of the given equation is,

 

${\mathrm{y}}_{\mathrm{c}}={\mathrm{c}}_1{\mathrm{e}}^{\mathrm{\theta }}+{\mathrm{c}}_2{\mathrm{e}}^{-\mathrm{\theta }}$

Step 3: Now find the particular solution to find a general solution for the equation.

Assume, the particular solution of equation (1),

 

${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{\theta }\right)=\mathrm{Asin\theta }+\mathrm{Bcos\theta }+{\mathrm{Ce}}^{2\mathrm{\theta }}......\left(2\right)$

 

Now find the first and second derivatives of the above equation,

 $\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\text{'}\left(\mathrm{\theta }\right)=\mathrm{Acos\theta }-\mathrm{Bsin\theta }+2{\mathrm{Ce}}^{2\mathrm{\theta }}\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{\theta }\right)=-\mathrm{Asin\theta }-\mathrm{Bcos\theta }+4{\mathrm{Ce}}^{2\mathrm{\theta }}\end{array}$

Substitute the value of  ${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{\theta }\right)$ and ${\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{\theta }\right)$ the equation (1),

$\begin{array}{c}\mathrm{y}\text{'}\text{'}\left(\mathrm{\theta }\right)-\mathrm{y}\left(\mathrm{\theta }\right)=\mathrm{sin\theta }-{\mathrm{e}}^{2\mathrm{\theta }}\\ -\mathrm{Asin\theta }-\mathrm{Bcos\theta }+4{\mathrm{Ce}}^{2\mathrm{\theta }}-[\mathrm{Asin\theta }+\mathrm{Bcos\theta }+{\mathrm{Ce}}^{2\mathrm{\theta }}]=\mathrm{sin\theta }-{\mathrm{e}}^{2\mathrm{\theta }}\\ -2\mathrm{Asin\theta }-2\mathrm{Bcos\theta }+3{\mathrm{Ce}}^{2\mathrm{\theta }}=\mathrm{sin\theta }-{\mathrm{e}}^{2\mathrm{\theta }}\end{array}$

Comparing all coefficients of the above equation,

  $\begin{array}{c}-2\mathrm{A}=1\Rightarrow \mathrm{A}=\frac{-1}{2}\\ -2\mathrm{B}=0\Rightarrow \mathrm{B}=0\\ 3\mathrm{C}=-1\Rightarrow \mathrm{C}=\frac{-1}{3}\end{array}$

Substitute the value of A, B, and C in the equation (2),

 $\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{\theta }\right)=-\frac{1}{2}\mathrm{sin\theta }+\left(0\right)\mathrm{cos\theta }-\frac{1}{3}{\mathrm{e}}^{2\mathrm{\theta }}\\ {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{\theta }\right)=-\frac{1}{2}\mathrm{sin\theta }-\frac{1}{3}{\mathrm{e}}^{2\mathrm{\theta }}\end{array}$

Step 4: The general solution and use the given initial condition.

Therefore, the general solution is,

$\begin{array}{c}\mathrm{y}={\mathrm{y}}_{\mathrm{c}}\left(\mathrm{\theta }\right)+{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{\theta }\right)\\ \mathrm{y}={\mathrm{c}}_1{\mathrm{e}}^{\mathrm{\theta }}+{\mathrm{c}}_2{\mathrm{e}}^{-\mathrm{\theta }}-\frac{1}{2}\mathrm{sin\theta }-\frac{1}{3}{\mathrm{e}}^{2\mathrm{\theta }}......\left(3\right)\end{array}$

Given the initial condition,

 

$\mathrm{y}\left(0\right)=1,\mathrm{y}\text{'}\left(0\right)=-1$

 

Substitute the value of y = 1 and  $\mathrm{\theta }=0$ in the equation (3),

$\begin{array}{c}\mathrm{y}={\mathrm{c}}_1{\mathrm{e}}^{\mathrm{\theta }}+{\mathrm{c}}_2{\mathrm{e}}^{-\mathrm{\theta }}-\frac{1}{2}\mathrm{sin\theta }-\frac{1}{3}{\mathrm{e}}^{2\mathrm{\theta }}\\ 1={\mathrm{c}}_1{\mathrm{e}}^0+{\mathrm{c}}_2{\mathrm{e}}^{-0}-\frac{1}{2}\sin \left(0\right)-\frac{1}{3}{\mathrm{e}}^{2\left(0\right)}\\ 1={\mathrm{c}}_1+{\mathrm{c}}_2-\frac{1}{3}\\ {\mathrm{c}}_1+{\mathrm{c}}_2=\frac{4}{3}......\left(4\right)\end{array}$

Now find the derivative of the equation (3),

$\mathrm{y}\text{'}={\mathrm{c}}_1{\mathrm{e}}^{\mathrm{\theta }}-{\mathrm{c}}_2{\mathrm{e}}^{-\mathrm{\theta }}-\frac{1}{2}\mathrm{cos\theta }-\frac{2}{3}{\mathrm{e}}^{2\mathrm{\theta }}$

Substitute the value of y’ = -1 and $\mathrm{\theta }=0$ in the above equation,

$\begin{array}{c}\mathrm{y}\text{'}={\mathrm{c}}_1{\mathrm{e}}^{\mathrm{\theta }}-{\mathrm{c}}_2{\mathrm{e}}^{-\mathrm{\theta }}-\frac{1}{2}\mathrm{cos\theta }-\frac{2}{3}{\mathrm{e}}^{2\mathrm{\theta }}\\ -1={\mathrm{c}}_1{\mathrm{e}}^0-{\mathrm{c}}_2{\mathrm{e}}^{-0}-\frac{1}{2}\cos \left(0\right)-\frac{2}{3}{\mathrm{e}}^{2\left(0\right)}\\ -1={\mathrm{c}}_1-{\mathrm{c}}_2-\frac{1}{2}-\frac{2}{3}\\ {\mathrm{c}}_1-{\mathrm{c}}_2=\frac{1}{6}.....\left(5\right)\end{array}$

Solve the equation (4) and (5),

$\begin{array}{l}{\mathrm{c}}_1+{\mathrm{c}}_2=\frac{4}{3}\\ {\mathrm{c}}_1-{\mathrm{c}}_2=\frac{1}{6}\\ \overline{\begin{array}{l}2{\mathrm{c}}_1=\frac{9}{6}\\ {\mathrm{c}}_1=\frac{3}{4}\end{array}}\end{array}$

Substitute the value of ${\mathrm{c}}_1$ in the equation (4),

$\begin{array}{c}{\mathrm{c}}_1+{\mathrm{c}}_2=\frac{4}{3}\\ \frac{3}{4}+{\mathrm{c}}_2=\frac{4}{3}\\ {\mathrm{c}}_2=\frac{7}{12}\end{array}$

Substitute the value of  ${\mathrm{c}}_1$and ${\mathrm{c}}_2$ in the equation (3),

$\begin{array}{c}\mathrm{y}={\mathrm{c}}_1{\mathrm{e}}^{\mathrm{\theta }}+{\mathrm{c}}_2{\mathrm{e}}^{-\mathrm{\theta }}-\frac{1}{2}\mathrm{sin\theta }-\frac{1}{3}{\mathrm{e}}^{2\mathrm{\theta }}\\ \mathrm{y}=\frac{3}{4}{\mathrm{e}}^{\mathrm{\theta }}+\frac{7}{12}{\mathrm{e}}^{-\mathrm{\theta }}-\frac{1}{2}\mathrm{sin\theta }-\frac{1}{3}{\mathrm{e}}^{2\mathrm{\theta }}\end{array}$

Thus, the solution to the initial value is:

$\mathrm{y}=\frac{3}{4}{\mathrm{e}}^{\mathrm{\theta }}+\frac{7}{12}{\mathrm{e}}^{-\mathrm{\theta }}-\frac{1}{2}\mathrm{sin\theta }-\frac{1}{3}{\mathrm{e}}^{2\mathrm{\theta }}$