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### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

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

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

See the step by step solution

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

The given differential equation is,

$\mathrm{y}\text{'}\text{'}\text{'}+\mathrm{y}\text{'}\text{'}-2\mathrm{y}={\mathrm{te}}^{\mathrm{t}}+1\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}......\left(1\right)$

Consider the particular solution is,

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=\mathrm{t}\left(\mathrm{At}+\mathrm{B}\right){\mathrm{e}}^{\mathrm{t}}+\mathrm{C}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}.....\left(2\right)\\ {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=\left({\mathrm{At}}^{2}+\mathrm{Bt}\right){\mathrm{e}}^{\mathrm{t}}+\mathrm{C}\end{array}$

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

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

Substitute value of ${\mathrm{y}}_{\mathrm{p}}\text{'}\left(\mathrm{t}\right),\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{t}\right)$ and ${\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\text{'}\left(\mathrm{t}\right)$ in the equation (1),

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

Comparing the all coefficients of the above equation;

role="math" localid="1655108362542" $\begin{array}{c}10\mathrm{A}=1\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}⇒\mathrm{A}=\frac{1}{10}\\ -2\mathrm{C}=1\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}⇒\mathrm{C}=\frac{-1}{2}\\ 8\mathrm{A}+5\mathrm{B}=0\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}.......\left(3\right)\end{array}$

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),

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