Q18E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problems 11–18, find a general solution to the differential equation.

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{6}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{9}\mathbf{y}\mathbf{=}{\mathbf{t}}^{\mathbf{-}\mathbf{3}}{\mathbf{e}}^{\mathbf{3}\mathbf{t}}$

Step-by-Step Solution

Answer

The general solution is $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_1{\mathrm{e}}^{3\mathrm{t}}+{\mathrm{c}}_2{\mathrm{te}}^{3\mathrm{t}}-\frac{1}{2}{\mathrm{t}}^{-1}{\mathrm{e}}^{3\mathrm{t}}$.

Step 1: Find a particular solution .

The differential equation is $\mathrm{y}\text{'}\text{'}-6\mathrm{y}\text{'}+9\mathrm{y}={\mathrm{t}}^{-3}{\mathrm{e}}^{3\mathrm{t}}$.

 

The homogenous equation is ${\mathrm{r}}^2-6\mathrm{r}+9=0$.

 

Two independent solutions are $\mathrm{r}=3,3$.

 

Then ${\mathrm{y}}_1={\mathrm{e}}^{3\mathrm{t}},{\mathrm{y}}_2={\mathrm{te}}^{3\mathrm{t}}$

${\mathrm{y}}_{\mathrm{h}}\left(\mathrm{t}\right)={\mathrm{c}}_1{\mathrm{e}}^{3\mathrm{t}}+{\mathrm{c}}_2{\mathrm{te}}^{3\mathrm{t}}$

The particular solution is ${\mathrm{y}}_{\mathrm{p}}={\mathrm{v}}_1\left(\mathrm{t}\right){\mathrm{e}}^{3\mathrm{t}}+{\mathrm{v}}_2\left(\mathrm{t}\right)\mathrm{t}{\mathrm{e}}^{3\mathrm{t}}$.

Step 2: Evaluate v 1   and   v 2

Here ${\mathrm{y}}_{\mathrm{p}}={\mathrm{v}}_1\left(\mathrm{t}\right){\mathrm{e}}^{3\mathrm{t}}+{\mathrm{v}}_2\left(\mathrm{t}\right)\mathrm{t}{\mathrm{e}}^{3\mathrm{t}}$

 

And referring to (9) and solve the system then

$$\begin{gathered} {\mathrm{v}}_1\text{'}{\mathrm{e}}^{3\mathrm{t}}+{\mathrm{v}}_2\text{'}{\mathrm{te}}^{3\mathrm{t}}=0 \\ {\mathrm{v}}_1\text{'}{\mathrm{e}}^{3\mathrm{t}}+\mathrm{v}{\text{'}}_2(\mathrm{t}{\mathrm{e}}^{3\mathrm{t}}+{\mathrm{e}}^{3\mathrm{t}})=\frac{\mathrm{f}}{\mathrm{a}} \\ {\mathrm{v}}_1\text{'}{\mathrm{e}}^{3\mathrm{t}}+\mathrm{v}{\text{'}}_2(\mathrm{t}{\mathrm{e}}^{3\mathrm{t}}+{\mathrm{e}}^{3\mathrm{t}})={\mathrm{t}}^{-3}{\mathrm{e}}^{3\mathrm{t}} \end{gathered}$$

Step 3: Find v 1 ' and v 1

$\begin{array}{rcl}{\mathrm{v}}_1\text{'}& =& \frac{-\mathrm{f}\left(\mathrm{t}\right){\mathrm{y}}_2\left(\mathrm{t}\right)}{\mathrm{a}\left[{\mathrm{y}}_1\left(\mathrm{t}\right)\mathrm{y}{\text{'}}_2\left(\mathrm{t}\right)-\mathrm{y}{\text{'}}_1\left(\mathrm{t}\right){\mathrm{y}}_2\left(\mathrm{t}\right)\right]}\\ & =& \frac{-({\mathrm{t}}^{-3}{\mathrm{e}}^{3\mathrm{t}}).{\mathrm{te}}^{3\mathrm{t}}}{\left[{\mathrm{e}}^{3\mathrm{t}}.(3\mathrm{t}{\mathrm{e}}^{3\mathrm{t}}+{\mathrm{e}}^{3\mathrm{t}})-3{\mathrm{e}}^{3\mathrm{t}}.{\mathrm{te}}^{3\mathrm{t}}\right]}\\ & =& \frac{-({\mathrm{t}}^{-3}{\mathrm{e}}^{3\mathrm{t}}).{\mathrm{te}}^{3\mathrm{t}}}{\left[{\mathrm{e}}^{6\mathrm{t}}\right]}\\ & =& {\mathrm{t}}^{-2}\end{array}$

Now integrating this.

$$\begin{gathered} {\mathrm{v}}_1\left(\mathrm{t}\right)=\int -{\mathrm{t}}^{-2}\mathrm{dt} \\ ={\mathrm{t}}^{-1} \end{gathered}$$ 

Step 4: Determine v 2 ' and v 2

$\begin{array}{rcl}{\mathrm{v}}_2\text{'}& =& \frac{\mathrm{f}\left(\mathrm{t}\right){\mathrm{y}}_1\left(\mathrm{t}\right)}{\mathrm{a}\left[{\mathrm{y}}_1\left(\mathrm{t}\right)\mathrm{y}{\text{'}}_2\left(\mathrm{t}\right)-\mathrm{y}{\text{'}}_1\left(\mathrm{t}\right){\mathrm{y}}_2\left(\mathrm{t}\right)\right]}\\ & =& \frac{{\mathrm{t}}^{-3}{\mathrm{e}}^{3\mathrm{t}}.{\mathrm{e}}^{3\mathrm{t}}}{\left[{\mathrm{e}}^{3\mathrm{t}}.(3\mathrm{t}{\mathrm{e}}^{3\mathrm{t}}+{\mathrm{e}}^{3\mathrm{t}})-3{\mathrm{e}}^{3\mathrm{t}}.{\mathrm{te}}^{3\mathrm{t}}\right]}\\ & =& \frac{\left({\mathrm{t}}^{-3}{\mathrm{e}}^{3\mathrm{t}}\right).{\mathrm{e}}^{3\mathrm{t}}}{\left[{\mathrm{e}}^{6\mathrm{t}}\right]}\\ & =& {\mathrm{t}}^{-3}\end{array}$ 

Integrate this.

$$\begin{gathered} {\mathrm{y}}_2\left(\mathrm{t}\right)=\int {\mathrm{t}}^{-3}\mathrm{dt} \\ =-\frac{{\mathrm{t}}^{-2}}{2} \end{gathered}$$

 

Thus, a particular solution is:

$$\begin{gathered} {\mathrm{y}}_{\mathrm{p}}=({\mathrm{t}}^{-1}){\mathrm{e}}^{3\mathrm{t}}+(-\frac{{\mathrm{t}}^{-2}}{2})\mathrm{t}{\mathrm{e}}^{3\mathrm{t}} \\ {\mathrm{y}}_{\mathrm{p}}=-\frac{1}{2}{\mathrm{t}}^{-1}{\mathrm{e}}^{3\mathrm{t}} \end{gathered}$$ 

And the general solution is: 

$$\begin{gathered} \mathrm{y}\left(\mathrm{t}\right)={\mathrm{y}}_{\mathrm{h}}\left(\mathrm{t}\right)+{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right) \\ \mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_1{\mathrm{e}}^{3\mathrm{t}}+{\mathrm{c}}_2{\mathrm{te}}^{3\mathrm{t}}-\frac{1}{2}{\mathrm{t}}^{-1}{\mathrm{e}}^{3\mathrm{t}} \end{gathered}$$

This is the required result.