Q28E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find the solution to the initial value problem.

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{\text{'}}\mathbf{-}\mathbf{12}\mathbf{y}\mathbf{=}{\mathbf{e}}^t\mathbf{+}{\mathbf{e}}^{2t}\mathbf{-}\mathbf{1}\mathbf{;}\mathbf{y}\left(\mathbf{0}\right)\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\left(\mathbf{0}\right)\mathbf{=}\mathbf{3}$

Step-by-Step Solution

Answer

The solution to the initial value problem is:

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

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

The differential equation is,

 

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

 

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,

 

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

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

The root of an auxiliary equation is, 

 

${\mathrm{m}}_1=3,{\mathrm{m}}_2=-4$

 

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),

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

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

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

Comparing all coefficients of the above equation,

$\begin{array}{c}-10\mathrm{A}=1\Rightarrow \mathrm{A}=\frac{-1}{10}\\ -6\mathrm{B}=1\Rightarrow \mathrm{B}=\frac{-1}{6}\\ -12\mathrm{C}=-1\Rightarrow \mathrm{C}=\frac{1}{12}\end{array}$

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,

 

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

 

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

$\begin{array}{c}\mathrm{y}={\mathrm{c}}_1{\mathrm{e}}^{3\mathrm{t}}+{\mathrm{c}}_2{\mathrm{e}}^{-4\mathrm{t}}-\frac{1}{10}{\mathrm{e}}^{\mathrm{t}}-\frac{1}{6}{\mathrm{e}}^{2\mathrm{t}}+\frac{1}{12}\\ 1={\mathrm{c}}_1{\mathrm{e}}^{3\left(0\right)}+{\mathrm{c}}_2{\mathrm{e}}^{-4\left(0\right)}-\frac{1}{10}{\mathrm{e}}^0-\frac{1}{6}{\mathrm{e}}^{2\left(0\right)}+\frac{1}{12}\\ 1={\mathrm{c}}_1+{\mathrm{c}}_2-\frac{1}{10}-\frac{1}{6}+\frac{1}{12}\\ {\mathrm{c}}_1+{\mathrm{c}}_2=\frac{71}{60}. .....\left(4\right)\end{array}$

Now find the derivative of the equation (3),

 

$\mathrm{y}\text{'}=3{\mathrm{c}}_1{\mathrm{e}}^{3\mathrm{t}}-4{\mathrm{c}}_2{\mathrm{e}}^{-4\mathrm{t}}-\frac{1}{10}{\mathrm{e}}^{\mathrm{t}}-\frac{1}{3}{\mathrm{e}}^{2\mathrm{t}}$

 

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

$\begin{array}{c}\mathrm{y}\text{'}=3{\mathrm{c}}_1{\mathrm{e}}^{3\mathrm{t}}-4{\mathrm{c}}_2{\mathrm{e}}^{-4\mathrm{t}}-\frac{1}{10}{\mathrm{e}}^{\mathrm{t}}-\frac{1}{3}{\mathrm{e}}^{2\mathrm{t}}\\ 3=3{\mathrm{c}}_1{\mathrm{e}}^{3\left(0\right)}-4{\mathrm{c}}_2{\mathrm{e}}^{-4\left(0\right)}-\frac{1}{10}{\mathrm{e}}^0-\frac{1}{3}{\mathrm{e}}^{2\left(0\right)}\\ 3=3{\mathrm{c}}_1-4{\mathrm{c}}_2-\frac{1}{10}-\frac{1}{3}\\ 3{\mathrm{c}}_1-4{\mathrm{c}}_2=3+\frac{13}{30}\\ 3{\mathrm{c}}_1-4{\mathrm{c}}_2=\frac{103}{30}. .....\left(5\right)\end{array}$

Solve the equation (4) and (5),

  $\begin{array}{c}4({\mathrm{c}}_1+{\mathrm{c}}_2)=\frac{71}{60}\times 4\\ 4{\mathrm{c}}_1+4{\mathrm{c}}_2=\frac{71}{15}\\ 3{\mathrm{c}}_1-4{\mathrm{c}}_2=\frac{103}{30}\\ {\mathrm{c}}_1=\frac{35}{30}\\ {\mathrm{c}}_1=\frac{7}{6}\end{array}$

Substitute the value of  ${\mathrm{c}}_1$in the equation (4),

$\begin{array}{c}{\mathrm{c}}_1+{\mathrm{c}}_2=\frac{71}{60}\\ \frac{7}{6}+{\mathrm{c}}_2=\frac{71}{60}\\ {\mathrm{c}}_2=\frac{71}{60}-\frac{7}{6}\\ {\mathrm{c}}_2=\frac{1}{60}\end{array}$

Substitute the value of  ${\mathrm{c}}_1$ and ${\mathrm{c}}_2$ in the equation (3),

 

$\begin{array}{c}\mathrm{y}={\mathrm{c}}_1{\mathrm{e}}^{3\mathrm{t}}+{\mathrm{c}}_2{\mathrm{e}}^{-4\mathrm{t}}-\frac{1}{10}{\mathrm{e}}^{\mathrm{t}}-\frac{1}{6}{\mathrm{e}}^{2\mathrm{t}}+\frac{1}{12}\\ \mathrm{y}=\frac{7}{6}{\mathrm{e}}^{3\mathrm{t}}+\frac{1}{60}{\mathrm{e}}^{-4\mathrm{t}}-\frac{1}{10}{\mathrm{e}}^{\mathrm{t}}-\frac{1}{6}{\mathrm{e}}^{2\mathrm{t}}+\frac{1}{12}\end{array}$

Thus, the solution to the initial value problem is:

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