Q24E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problems 22 through 25, use variation of parameters to find a general solution to the differential equation given that the functions ${\mathbf{y}}_{\mathbf{1}}$ and ${\mathbf{y}}_{\mathbf{2}}$ are linearly independent solutions to the corresponding homogeneous equation for t> 0. $\mathbf{ty}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{(}\mathbf{1}\mathbf{-}\mathbf{2}\mathbf{t}\mathbf{)}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{(}\mathbf{t}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{y}\mathbf{=}\mathbf{t}{\mathbf{e}}^t\mathbf{;}{\mathbf{y}}_1\mathbf{=}{\mathbf{e}}^t\mathbf{,}{\mathbf{y}}_2\mathbf{=}{\mathbf{e}}^t\mathbf{lnt}$

Step-by-Step Solution

Answer

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

Step 1: Find particular solution.

Given the differential equation is $\mathrm{ty}\text{'}\text{'}+(1-2\mathrm{t})\mathrm{y}\text{'}+(\mathrm{t}-1)\mathrm{y}=\mathrm{t}{\mathrm{e}}^{\mathrm{t}}$

 

And ${\mathrm{y}}_1={\mathrm{e}}^{\mathrm{t}},{\mathrm{y}}_2={\mathrm{e}}^{\mathrm{t}}\mathrm{lnt}$

 

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

 

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

Step 2: Evaluate v 1    and    v 2 .

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

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

 

Now integrate the above result.

$\begin{array}{c}{\mathrm{v}}_1\left(\mathrm{t}\right)=\int -\mathrm{t}\left(\mathrm{lnt}\right)\mathrm{dt}\\ =-(\mathrm{lnt}+1)\end{array}$

 

Step 3: Determine v 2 ' and v 2

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

Integrate the above result, we get:

$\begin{array}{c}{\mathrm{v}}_2\left(\mathrm{t}\right)=\int \mathrm{tdt}\\ =\frac{{\mathrm{t}}^2}{2}\end{array}$

Therefore the particular solution is:

 ${\mathrm{y}}_{\mathrm{p}}=-(\mathrm{lnt}+1){\mathrm{e}}^{\mathrm{t}}+\frac{{\mathrm{t}}^2}{2}{\mathrm{e}}^{\mathrm{t}}\mathrm{lnt}$

Thus, the general solution is:

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