Q30E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find the solution to the initial value problem. $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{=}{\mathbf{t}}^2\mathbf{+}\mathbf{1}\mathbf{-}{\mathbf{e}}^t\mathbf{y}\left(\mathbf{0}\right)\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\left(\mathbf{0}\right)\mathbf{=}\mathbf{2}$

Step-by-Step Solution

Answer

The solution to the initial value problem is given as:

$\mathrm{y}=-\frac{27}{4}{\mathrm{e}}^{-\mathrm{t}}-\frac{1}{2}{\mathrm{te}}^{-\mathrm{t}}+{\mathrm{t}}^2-4\mathrm{t}+7-\frac{1}{4}{\mathrm{e}}^{\mathrm{t}}$.

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

The differential equation is,

$\mathrm{y}\text{'}\text{'}+2\mathrm{y}\text{'}+\mathrm{y}={\mathrm{t}}^2+1-{\mathrm{e}}^{\mathrm{t}}\mathrm{......}\left(\mathbf{1}\right)$

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

$\mathrm{y}\text{'}\text{'}+2\mathrm{y}\text{'}+\mathrm{y}=0$

The auxiliary equation for the above equation,

$\begin{array}{c}{\mathrm{m}}^2+2\mathrm{m}+1=0\\ {(\mathrm{m}+1)}^2=0\end{array}$

Step 2: Now 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,

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

 

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

Assume, the particular solution of equation (1),

 ${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{At}}^2+\mathrm{Bt}+\mathrm{C}+{\mathrm{De}}^{\mathrm{t}}.....\left(2\right)$

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

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\text{'}\left(\mathrm{t}\right)=2\mathrm{At}+\mathrm{B}+{\mathrm{De}}^{\mathrm{t}}\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{t}\right)=2\mathrm{A}+{\mathrm{De}}^{\mathrm{t}}\end{array}$

Substitute the value of  ${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right),{\mathrm{y}}_{\mathrm{p}}\text{'}\left(\mathrm{t}\right)$ and  ${\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{t}\right)$ the equation (1),

$\begin{array}{c}\mathrm{y}\text{'}\text{'}+2\mathrm{y}\text{'}+\mathrm{y}={\mathrm{t}}^2+1-{\mathrm{e}}^{\mathrm{t}}\\ 2\mathrm{A}+{\mathrm{De}}^{\mathrm{t}}+2[2\mathrm{At}+\mathrm{B}+{\mathrm{De}}^{\mathrm{t}}]+{\mathrm{At}}^2+\mathrm{Bt}+\mathrm{C}+{\mathrm{De}}^{\mathrm{t}}={\mathrm{t}}^2+1-{\mathrm{e}}^{\mathrm{t}}\\ {\mathrm{At}}^2+(\mathrm{B}+4\mathrm{A})\mathrm{t}+(2\mathrm{A}+2\mathrm{B}+\mathrm{C})+4{\mathrm{De}}^{\mathrm{t}}={\mathrm{t}}^2+1-{\mathrm{e}}^{\mathrm{t}}\end{array}$

Comparing all coefficients of the above equation,

$\begin{array}{c}4\mathrm{D}=-1\Rightarrow \mathrm{D}=\frac{-1}{4}\\ \mathrm{A}=1\\ \mathrm{B}+4\mathrm{A}=0......\left(3\right)\\ 2\mathrm{A}+2\mathrm{B}+\mathrm{C}=1......\left(4\right)\end{array}$

Substitute the value of A in the equation (3),

$\begin{array}{c}\mathrm{B}+4\left(1\right)=0\\ \mathrm{B}=-4\end{array}$

Substitute the value of A and B in the equation (4),

$\begin{array}{c}2\left(1\right)+2(-4)+\mathrm{C}=1\\ \mathrm{C}=7\end{array}$

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

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{At}}^2+\mathrm{Bt}+\mathrm{C}+{\mathrm{De}}^{\mathrm{t}}\\ {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{t}}^2-4\mathrm{t}+7-\frac{1}{4}{\mathrm{e}}^{\mathrm{t}}\end{array}$

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

Therefore, the general solution is,

$\begin{array}{l}\mathrm{y}={\mathrm{y}}_{\mathrm{c}}\left(\mathrm{t}\right)+{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)\\ \mathrm{y}={\mathrm{c}}_1{\mathrm{e}}^{-\mathrm{t}}+{\mathrm{c}}_2{\mathrm{te}}^{-\mathrm{t}}+{\mathrm{t}}^2-4\mathrm{t}+7-\frac{1}{4}{\mathrm{e}}^{\mathrm{t}}. .....\left(5\right)\end{array}$

Given the initial condition,

$\mathrm{y}\left(0\right)=0,\mathrm{y}\text{'}\left(0\right)=2$

Substitute the value of y = 0 and t = 0 in the equation (5),

$\begin{array}{c}\mathrm{y}={\mathrm{c}}_1{\mathrm{e}}^{-\mathrm{t}}+{\mathrm{c}}_2{\mathrm{te}}^{-\mathrm{t}}+{\mathrm{t}}^2-4\mathrm{t}+7-\frac{1}{4}{\mathrm{e}}^{\mathrm{t}}\\ 0={\mathrm{c}}_1{\mathrm{e}}^{-0}+{\mathrm{c}}_2\left(0\right){\mathrm{e}}^{-0}+{\left(0\right)}^2-4\left(0\right)+7-\frac{1}{4}{\mathrm{e}}^{\left(0\right)}\\ 0={\mathrm{c}}_1+7-\frac{1}{4}\\ {\mathrm{c}}_1=-\frac{27}{4}\end{array}$

Now find the derivative of the equation (5),

$\begin{array}{c}\mathrm{y}\text{'}=-{\mathrm{c}}_1{\mathrm{e}}^{-\mathrm{t}}-{\mathrm{c}}_2{\mathrm{te}}^{-\mathrm{t}}+{\mathrm{c}}_2{\mathrm{e}}^{-\mathrm{t}}+2\mathrm{t}-4-\frac{1}{4}{\mathrm{e}}^{\mathrm{t}}\\ \mathrm{y}\text{'}=-{\mathrm{c}}_1{\mathrm{e}}^{-\mathrm{t}}+{\mathrm{c}}_2({\mathrm{e}}^{-\mathrm{t}}-{\mathrm{te}}^{-\mathrm{t}})+2\mathrm{t}-4-\frac{1}{4}{\mathrm{e}}^{\mathrm{t}}\end{array}$

Substitute the value of y’ = 2 and t = 0 in the above equation,

$\begin{array}{c}\mathrm{y}\text{'}=-{\mathrm{c}}_1{\mathrm{e}}^{-\mathrm{t}}+{\mathrm{c}}_2({\mathrm{e}}^{-\mathrm{t}}-{\mathrm{te}}^{-\mathrm{t}})+2\mathrm{t}-4-\frac{1}{4}{\mathrm{e}}^{\mathrm{t}}\\ 2=-{\mathrm{c}}_1{\mathrm{e}}^{-0}+{\mathrm{c}}_2({\mathrm{e}}^{-0}-\left(0\right){\mathrm{e}}^{-0})+2\left(0\right)-4-\frac{1}{4}{\mathrm{e}}^0\\ 2=-{\mathrm{c}}_1+{\mathrm{c}}_2-4-\frac{1}{4}\\ -{\mathrm{c}}_1+{\mathrm{c}}_2=\frac{25}{4}. ....\left(6\right)\end{array}$

Substitute the value of  ${\mathbf{c}}_{\mathbf{1}}$  in the equation (6),

$\begin{array}{c}-\left(\frac{-27}{4}\right)+{\mathrm{c}}_2=\frac{25}{4}\\ {\mathrm{c}}_2=\frac{25}{4}-\frac{27}{4}\\ {\mathrm{c}}_2=-\frac{1}{2}\end{array}$

Substitute the value of  ${\mathrm{c}}_1$ and ${\mathrm{c}}_2$ in the equation (5),

$$\begin{gathered} \mathrm{y}={\mathrm{c}}_1{\mathrm{e}}^{-\mathrm{t}}+{\mathrm{c}}_2{\mathrm{te}}^{-\mathrm{t}}+{\mathrm{t}}^2-4\mathrm{t}+7-\frac{1}{4}{\mathrm{e}}^{\mathrm{t}} \\ \mathrm{y}=-\frac{27}{4}{\mathrm{e}}^{-\mathrm{t}}-\frac{1}{2}{\mathrm{te}}^{-\mathrm{t}}+{\mathrm{t}}^2-4\mathrm{t}+7-\frac{1}{4}{\mathrm{e}}^{\mathrm{t}} \end{gathered}$$