Q39E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find a particular solution to the given higher-order equation. $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{2}\mathbf{y}\mathbf{=}{\mathbf{te}}^t\mathbf{+}\mathbf{1}$

Step-by-Step Solution

Answer

Thus, the particular solution is  ${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=\mathrm{t}(\frac{1}{10}\mathrm{t}-\frac{4}{25}){\mathrm{e}}^{\mathrm{t}}-\frac{1}{2}.$          

Step 1: Consider the particular solution for the given differential equation.

The given differential equation is,

 

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

 

Consider the particular solution is,

 

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=\mathrm{t}(\mathrm{At}+\mathrm{B}){\mathrm{e}}^{\mathrm{t}}+\mathrm{C}. ....\left(2\right)\\ {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=({\mathrm{At}}^2+\mathrm{Bt}){\mathrm{e}}^{\mathrm{t}}+\mathrm{C}\end{array}$

 

Take first, second and third derivative of the above equation,

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\text{'}\left(\mathrm{t}\right)={\mathrm{e}}^{\mathrm{t}}(2\mathrm{At}+\mathrm{B})+{\mathrm{e}}^{\mathrm{t}}({\mathrm{At}}^2+\mathrm{Bt})\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\left(\mathrm{t}\right)={\mathrm{e}}^{\mathrm{t}}({\mathrm{At}}^2+(2\mathrm{A}+\mathrm{B})\mathrm{t}+\mathrm{B})\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{t}\right)={\mathrm{e}}^{\mathrm{t}}({\mathrm{At}}^2+(4\mathrm{A}+\mathrm{B})\mathrm{t}+2\mathrm{A}+2\mathrm{B})\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\text{'}\left(\mathrm{t}\right)={\mathrm{e}}^{\mathrm{t}}({\mathrm{At}}^2+(6\mathrm{A}+\mathrm{B})\mathrm{t}+6\mathrm{A}+3\mathrm{B})\end{array}$

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

$\begin{array}{c}\mathrm{y}\text{'}\text{'}\text{'}+\mathrm{y}\text{'}\text{'}-2\mathrm{y}={\mathrm{te}}^{\mathrm{t}}+1\\ {\mathrm{e}}^{\mathrm{t}}({\mathrm{At}}^2+(6\mathrm{A}+\mathrm{B})\mathrm{t}+6\mathrm{A}+3\mathrm{B})+{\mathrm{e}}^{\mathrm{t}}({\mathrm{At}}^2+(4\mathrm{A}+\mathrm{B})\mathrm{t}+2\mathrm{A}+2\mathrm{B})-2[{\mathrm{e}}^{\mathrm{t}}({\mathrm{At}}^2+\mathrm{Bt})+\mathrm{C}]={\mathrm{te}}^{\mathrm{t}}+1\\ {\mathrm{e}}^{\mathrm{t}}(8\mathrm{A}+5\mathrm{B})+{\mathrm{te}}^{\mathrm{t}}\left(10\mathrm{A}\right)-2\mathrm{C}={\mathrm{te}}^{\mathrm{t}}+1\end{array}$

Comparing the all coefficients of the above equation;

$\begin{array}{c}10\mathrm{A}=1\Rightarrow \mathrm{A}=\frac{1}{10}\\ -2\mathrm{C}=1\Rightarrow \mathrm{C}=\frac{-1}{2}\\ 8\mathrm{A}+5\mathrm{B}=0. ......\left(3\right)\end{array}$

Substitute the value A in the equation (3),

$\begin{array}{c}8\left(\frac{1}{10}\right)+5\mathrm{B}=0\\ \mathrm{B}=\frac{-4}{25}\end{array}$

Step 2: Conclusion.

Therefore, the particular solution of the equation (1),

${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=\mathrm{t}\left(\frac{1}{10}\mathrm{t}-\frac{4}{25}\right){\mathrm{e}}^{\mathrm{t}}-\frac{1}{2}$