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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

# Find a particular solution to the differential equation.${\mathbf{x}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}\left(\mathbf{t}\right){\mathbf{-}}{\mathbf{4}}{\mathbf{x}}{\mathbf{\text{'}}}\left(\mathbf{t}\right){\mathbf{+}}{\mathbf{4}}{\mathbf{x}}\left(\mathbf{t}\right){\mathbf{=}}{{\mathbf{te}}}^{2t}$

The particular solution is ${\mathbf{x}}_{p}\mathbf{=}\frac{1}{6}{\mathbf{t}}^{3}{\mathbf{e}}^{2t}.$

See the step by step solution

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

Consider the given differential equation,

$\mathrm{x}\text{'}\text{'}\left(\mathrm{t}\right)-4\mathrm{x}\text{'}\left(\mathrm{t}\right)+4\mathrm{x}\left(\mathrm{t}\right)={\mathrm{te}}^{2\mathrm{t}}\text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\dots \left(1\right)$

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

$\mathrm{x}\text{'}\text{'}\left(\mathrm{t}\right)-4\mathrm{x}\text{'}\left(\mathrm{t}\right)+4\mathrm{x}\left(\mathrm{t}\right)=0$

The auxiliary equation for the above equation,

${\mathrm{m}}^{2}-4\mathrm{m}+4=0$

## Step 2: Now find the roots of the auxiliary equation

Solve the auxiliary equation,

$\begin{array}{c}{\mathrm{m}}^{2}-4\mathrm{m}+4=0\\ {\left(\mathrm{m}-2\right)}^{2}=0\end{array}$

The roots of the auxiliary equation are;

${\mathrm{m}}_{1}=2,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}&\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{m}}_{2}=2$

The complementary solution of the given equation is;

${\mathrm{x}}_{\mathrm{c}}={\mathrm{c}}_{1}{\mathrm{e}}^{2\mathrm{t}}+{\mathrm{c}}_{2}{\mathrm{te}}^{2\mathrm{t}}$

## Step 2: Find a particular solution.

Therefore, the particular solution of equation (1),

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

Now find the derivative of the above equation,

$\begin{array}{c}{\mathrm{x}}_{\mathrm{p}}\text{'}\left(\mathrm{t}\right)=\left(3{\mathrm{At}}^{2}+2\mathrm{Bt}\right){\mathrm{e}}^{2\mathrm{t}}+2{\mathrm{t}}^{2}\left(\mathrm{At}+\mathrm{B}\right){\mathrm{e}}^{2\mathrm{t}}\\ {\mathrm{x}}_{\mathrm{p}}\text{'}\left(\mathrm{t}\right)=\left(3{\mathrm{At}}^{2}+2\mathrm{Bt}+2{\mathrm{At}}^{3}+2{\mathrm{t}}^{2}\mathrm{B}\right){\mathrm{e}}^{2\mathrm{t}}\\ {\mathrm{x}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{t}\right)=\left(12{\mathrm{At}}^{2}+8\mathrm{Bt}+6\mathrm{At}+4{\mathrm{At}}^{3}+4{\mathrm{t}}^{2}\mathrm{B}+2\mathrm{B}\right){\mathrm{e}}^{2\mathrm{t}}\end{array}$

From the equation (1), substitute the value of ${\mathrm{x}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{t}\right),\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}_{\mathrm{p}}\text{'}\left(\mathrm{t}\right)$ and ${\mathrm{x}}_{\mathrm{p}}\left(\mathrm{t}\right)$, we get

$\begin{array}{c}\mathrm{x}\text{'}{\text{'}}_{\mathrm{p}}\left(\mathrm{t}\right)-4{\mathrm{x}}_{\mathrm{p}}\text{'}\left(\mathrm{t}\right)+4{\mathrm{x}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{te}}^{2\mathrm{t}}\\ \left(12{\mathrm{At}}^{2}+8\mathrm{Bt}+6\mathrm{At}+4{\mathrm{At}}^{3}+4{\mathrm{t}}^{2}\mathrm{B}+2\mathrm{B}\right){\mathrm{e}}^{2\mathrm{t}}-4\left(3{\mathrm{At}}^{2}+2\mathrm{Bt}+2{\mathrm{At}}^{3}+2{\mathrm{t}}^{2}\mathrm{B}\right){\mathrm{e}}^{2\mathrm{t}}+4{\mathrm{t}}^{2}\left(\mathrm{At}+\mathrm{B}\right){\mathrm{e}}^{2\mathrm{t}}={\mathrm{te}}^{2\mathrm{t}}\\ {\mathrm{te}}^{2\mathrm{t}}\left(6\mathrm{A}\right)+2{\mathrm{Be}}^{2\mathrm{t}}={\mathrm{te}}^{2\mathrm{t}}\end{array}$

## Step 4: Final conclusion.

Comparing all coefficients of the above equation,

$\begin{array}{c}6\mathrm{A}=1\text{\hspace{0.17em}\hspace{0.17em}}⇒\mathrm{A}=\frac{1}{6}\\ \mathrm{B}=0\end{array}$

Therefore, the particular solution of equation (1),

$\begin{array}{c}{\mathrm{x}}_{\mathrm{p}}={\mathrm{t}}^{2}\left(\mathrm{At}+\mathrm{B}\right){\mathrm{e}}^{2\mathrm{t}}\\ {\mathrm{x}}_{\mathrm{p}}={\mathrm{t}}^{2}\left(\frac{}{}\mathrm{t}+0\right){\mathrm{e}}^{2\mathrm{t}}\\ {\mathrm{x}}_{\mathrm{p}}=\frac{}{}{\mathrm{t}}^{3}{\mathrm{e}}^{2\mathrm{t}}\end{array}$