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Found in: Page 186

### 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 the solution to the initial value problem.${\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}{\mathbf{+}}{\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{-}}{\mathbf{12}}{\mathbf{y}}{\mathbf{=}}{{\mathbf{e}}}^{t}{\mathbf{+}}{{\mathbf{e}}}^{2t}{\mathbf{-}}{\mathbf{1}}{\mathbf{;}}{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}}{\mathbf{y}}\left(\mathbf{0}\right){\mathbf{=}}{\mathbf{1}}{\mathbf{,}}{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}}{\mathbf{y}}{\mathbf{\text{'}}}\left(\mathbf{0}\right){\mathbf{=}}{\mathbf{3}}$

The solution to the initial value problem is:

$\mathbf{y}\mathbf{=}\frac{7}{6}{\mathbf{e}}^{3t}\mathbf{+}\frac{1}{60}{\mathbf{e}}^{-4t}\mathbf{-}\frac{1}{10}{\mathbf{e}}^{t}\mathbf{-}\frac{1}{6}{\mathbf{e}}^{2t}\mathbf{+}\frac{1}{12}$

See the step by step solution

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

The differential equation is,

$\mathrm{y}\text{'}\text{'}+\mathrm{y}\text{'}-12\mathrm{y}={\mathrm{e}}^{\mathrm{t}}+{\mathrm{e}}^{2\mathrm{t}}-1\text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}......\left(1\right)$

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

$\mathrm{y}\text{'}\text{'}+\mathrm{y}\text{'}-12\mathrm{y}=0$

The auxiliary equation for the above equation,

$\begin{array}{c}{\mathrm{m}}^{2}+\mathrm{m}-12=0\\ {\mathrm{m}}^{2}+4\mathrm{m}-3\mathrm{m}-12=0\\ \mathrm{m}\left(\mathrm{m}+4\right)-3\left(\mathrm{m}+4\right)=0\\ \left(\mathrm{m}-3\right)\left(\mathrm{m}+4\right)=0\end{array}$

## Step 2: Find the complementary solution of the given equation.

The root of an auxiliary equation is,

${\mathrm{m}}_{1}=3,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{m}}_{2}=-4$

The complementary solution of the given equation is,

${\mathrm{y}}_{\mathrm{c}}={\mathrm{c}}_{1}{\mathrm{e}}^{3\mathrm{t}}+{\mathrm{c}}_{2}{\mathrm{e}}^{-4\mathrm{t}}$

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

Assume, the particular solution of equation (1),

${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{Ae}}^{\mathrm{t}}+{\mathrm{Be}}^{2\mathrm{t}}+\mathrm{C}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}.....\left(2\right)$

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

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\text{'}\left(\mathrm{t}\right)={\mathrm{Ae}}^{\mathrm{t}}+2{\mathrm{Be}}^{2\mathrm{t}}\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{t}\right)={\mathrm{Ae}}^{\mathrm{t}}+4{\mathrm{Be}}^{2\mathrm{t}}\end{array}$

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

$\begin{array}{c}\mathrm{y}\text{'}\text{'}+\mathrm{y}\text{'}-12\mathrm{y}={\mathrm{e}}^{\mathrm{t}}+{\mathrm{e}}^{2\mathrm{t}}-1\\ {\mathrm{Ae}}^{\mathrm{t}}+4{\mathrm{Be}}^{2\mathrm{t}}+{\mathrm{Ae}}^{\mathrm{t}}+2{\mathrm{Be}}^{2\mathrm{t}}-12\left[{\mathrm{Ae}}^{\mathrm{t}}+{\mathrm{Be}}^{2\mathrm{t}}+\mathrm{C}\right]={\mathrm{e}}^{\mathrm{t}}+{\mathrm{e}}^{2\mathrm{t}}-1\\ -10{\mathrm{Ae}}^{\mathrm{t}}-6{\mathrm{Be}}^{2\mathrm{t}}-12\mathrm{C}={\mathrm{e}}^{\mathrm{t}}+{\mathrm{e}}^{2\mathrm{t}}-1\end{array}$

Comparing all coefficients of the above equation,

role="math" localid="1655098651285" $\begin{array}{c}-10\mathrm{A}=1\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}⇒\mathrm{A}=\frac{-1}{10}\\ -6\mathrm{B}=1\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}⇒\mathrm{B}=\frac{-1}{6}\\ -12\mathrm{C}=-1⇒\mathrm{C}=\frac{1}{12}\end{array}$

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

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{Ae}}^{\mathrm{t}}+{\mathrm{Be}}^{2\mathrm{t}}+\mathrm{C}\\ {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=-\frac{1}{10}{\mathrm{e}}^{\mathrm{t}}-\frac{1}{6}{\mathrm{e}}^{2\mathrm{t}}+\frac{1}{12}\end{array}$

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

Therefore, the general solution is,

$\begin{array}{c}\mathrm{y}={\mathrm{y}}_{\mathrm{c}}\left(\mathrm{t}\right)+{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)\\ \mathrm{y}={\mathrm{c}}_{1}{\mathrm{e}}^{3\mathrm{t}}+{\mathrm{c}}_{2}{\mathrm{e}}^{-4\mathrm{t}}-\frac{1}{10}{\mathrm{e}}^{\mathrm{t}}-\frac{1}{6}{\mathrm{e}}^{2\mathrm{t}}+\frac{1}{12}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} .}....\left(3\right)\end{array}$

Given the initial condition,

$\mathrm{y}\left(0\right)=1,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{y}\text{'}\left(0\right)=3$

Substitute the value of y = 1 and t = 0 in the equation (3),

role="math" localid="1655099043521" $\begin{array}{c}\mathrm{y}={\mathrm{c}}_{1}{\mathrm{e}}^{3\mathrm{t}}+{\mathrm{c}}_{2}{\mathrm{e}}^{-4\mathrm{t}}-\frac{1}{10}{\mathrm{e}}^{\mathrm{t}}-\frac{1}{6}{\mathrm{e}}^{2\mathrm{t}}+\frac{1}{12}\\ 1={\mathrm{c}}_{1}{\mathrm{e}}^{3\left(0\right)}+{\mathrm{c}}_{2}{\mathrm{e}}^{-4\left(0\right)}-\frac{1}{10}{\mathrm{e}}^{0}-\frac{1}{6}{\mathrm{e}}^{2\left(0\right)}+\frac{1}{12}\\ 1={\mathrm{c}}_{1}+{\mathrm{c}}_{2}-\frac{1}{10}-\frac{1}{6}+\frac{1}{12}\\ {\mathrm{c}}_{1}+{\mathrm{c}}_{2}=\frac{71}{60}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}......\left(4\right)\end{array}$

Now find the derivative of the equation (3),

role="math" localid="1655099101150" $\mathrm{y}\text{'}=3{\mathrm{c}}_{1}{\mathrm{e}}^{3\mathrm{t}}-4{\mathrm{c}}_{2}{\mathrm{e}}^{-4\mathrm{t}}-\frac{1}{10}{\mathrm{e}}^{\mathrm{t}}-\frac{1}{3}{\mathrm{e}}^{2\mathrm{t}}$

Substitute the value of y’ = 3 and t = 0 in the above equation,

role="math" localid="1655099214255" $\begin{array}{c}\mathrm{y}\text{'}=3{\mathrm{c}}_{1}{\mathrm{e}}^{3\mathrm{t}}-4{\mathrm{c}}_{2}{\mathrm{e}}^{-4\mathrm{t}}-\frac{1}{10}{\mathrm{e}}^{\mathrm{t}}-\frac{1}{3}{\mathrm{e}}^{2\mathrm{t}}\\ 3=3{\mathrm{c}}_{1}{\mathrm{e}}^{3\left(0\right)}-4{\mathrm{c}}_{2}{\mathrm{e}}^{-4\left(0\right)}-\frac{1}{10}{\mathrm{e}}^{0}-\frac{1}{3}{\mathrm{e}}^{2\left(0\right)}\\ 3=3{\mathrm{c}}_{1}-4{\mathrm{c}}_{2}-\frac{1}{10}-\frac{1}{3}\\ 3{\mathrm{c}}_{1}-4{\mathrm{c}}_{2}=3+\frac{13}{30}\\ 3{\mathrm{c}}_{1}-4{\mathrm{c}}_{2}=\frac{103}{30}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}......\left(5\right)\end{array}$

Solve the equation (4) and (5),

role="math" localid="1655099327421"

Substitute the value of ${\mathrm{c}}_{1}$in the equation (4),

role="math" localid="1655099442348" $\begin{array}{c}{\mathrm{c}}_{1}+{\mathrm{c}}_{2}=\frac{71}{60}\\ \frac{7}{6}+{\mathrm{c}}_{2}=\frac{71}{60}\\ {\mathrm{c}}_{2}=\frac{71}{60}-\frac{7}{6}\\ {\mathrm{c}}_{2}=\frac{1}{60}\end{array}$

Substitute the value of ${\mathrm{c}}_{1}$ and ${\mathrm{c}}_{2}$ in the equation (3),

role="math" localid="1655099619782" $\begin{array}{c}\mathrm{y}={\mathrm{c}}_{1}{\mathrm{e}}^{3\mathrm{t}}+{\mathrm{c}}_{2}{\mathrm{e}}^{-4\mathrm{t}}-\frac{1}{10}{\mathrm{e}}^{\mathrm{t}}-\frac{1}{6}{\mathrm{e}}^{2\mathrm{t}}+\frac{1}{12}\\ \mathrm{y}=\frac{7}{6}{\mathrm{e}}^{3\mathrm{t}}+\frac{1}{60}{\mathrm{e}}^{-4\mathrm{t}}-\frac{1}{10}{\mathrm{e}}^{\mathrm{t}}-\frac{1}{6}{\mathrm{e}}^{2\mathrm{t}}+\frac{1}{12}\end{array}$

Thus, the solution to the initial value problem is:

role="math" localid="1655099676357" $\mathrm{y}=\frac{7}{6}{\mathrm{e}}^{3\mathrm{t}}+\frac{1}{60}{\mathrm{e}}^{-4\mathrm{t}}-\frac{1}{10}{\mathrm{e}}^{\mathrm{t}}-\frac{1}{6}{\mathrm{e}}^{2\mathrm{t}}+\frac{1}{12}$