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Expert-verified Found in: Page 180 ### 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 # In Problems 35, use the method of undetermined coefficients to find a particular solution to the given higher-order equation.${\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}{\mathbf{+}}{\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}{\mathbf{-}}{\mathbf{2}}{\mathbf{y}}{\mathbf{=}}{{\mathbf{te}}}^{t}$

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

See the step by step solution

## Step 1: Firstly, write the auxiliary equation of the given differential equation.

The given differential equation is:

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

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

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

The auxiliary equation for the above equation,

${\mathrm{m}}^{3}+{\mathrm{m}}^{2}-2=0$

Solve the auxiliary equation,

role="math" localid="1654925783487" $\begin{array}{c}\left(\mathrm{m}-1\right)\left({\mathrm{m}}^{2}+2\mathrm{m}+2\right)=0\\ \mathrm{m}=1,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{m}=\frac{-2±\sqrt{4-8}}{2}\\ \mathrm{m}=1,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{m}=-1±\mathrm{i}\end{array}$

## Step 2: Use the method of undetermined coefficients to find a particular solution to a given differential equation.

Consider the particular solution is,

${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=\mathrm{t}\left(\mathrm{At}+\mathrm{B}\right){\mathrm{e}}^{\mathrm{t}}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\dots \left(2\right)$

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

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

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

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

Comparing the coefficients of the above equation;

role="math" localid="1654925973361" $\begin{array}{c}10\mathrm{A}=1\\ \mathrm{A}=\frac{1}{10}\\ 8\mathrm{A}+5\mathrm{B}=0\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(3\right)\end{array}$

Substitute the value A in the equation (3),

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

## Step 3: Conclusion.

Substitute values A and B in the equation (2),

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=\mathrm{t}\left(\mathrm{At}+\mathrm{B}\right){\mathrm{e}}^{\mathrm{t}}\\ {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=\mathrm{t}\left(\frac{1}{10}\mathrm{t}-\frac{4}{25}\right){\mathrm{e}}^{\mathrm{t}}\end{array}$

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

${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=\mathrm{t}\left(\frac{1}{10}\mathrm{t}-\frac{4}{25}\right){\mathrm{e}}^{\mathrm{t}}$ ### Want to see more solutions like these? 