Q3E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problems 1–8, find a general solution to the differential equation using the method of variation of parameters.

$\mathbf{3}\mathbf{.}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{2}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathit{y}\mathbf{=}{\mathit{t}}^{\mathbf{-}\mathbf{1}}{\mathit{e}}^{\mathbf{t}}$

Step-by-Step Solution

Answer

The general solution is $y\left(t\right)=c_1{e}^t+c_{2}t{e}^t-t{e}^t+ln\left|t\right|t{e}^t$.

Step 1: Find the particular solution .

The homogenous equation is $r^2-2r+1=0$.

 

Two independent solutions are $r=1,1$.

 

Then $y_1=e^t,y_2=t{e}^t$

$y_h\left(t\right)=c_1{e}^t+c_{2}t{e}^t$

 

The particular solution is $y_p=v_1\left(t\right)e^t+v_2\left(t\right)t{e}^t$

Step 2: Evaluate v 1     and     v 2 .

Here $y_p=v_1\left(t\right)e^t+v_2\left(t\right)t{e}^t$

 

And referring to (9) $y\left(t\right)=c_1{e}^{\alpha t}\cos \beta t{c}_2{e}^{\alpha t}\sin \beta t,$ and solve the system then

 

Put the value of $y_p=0$

$$\begin{gathered} v_1\text{'}{e}^t+v_2\text{'}t{e}^t=0 \\ v_1\text{'}{e}^t+v_2\text{'}\left[e^t+t{e}^t\right]=\frac{f}{a} \\ -v_1\text{'}{e}^t+v_2\text{'}\left[e^t+t{e}^t\right]=t^{-1}{e}^t \\ -v_1\text{'}+v_2\text{'}\left[1+t\right]=t^{-1} \end{gathered}$$ 

Step 3: Find v ' 1 and v 1 .

$$\begin{gathered} v_1\text{'}=\frac{-f\left(t\right)y_2\left(t\right)}{a\left[y_1\left(t\right)y{\text{'}}_2\left(t\right)-y{\text{'}}_1\left(t\right)y_2\left(t\right)\right]} \\ =\frac{-t^{-1}{e}^t.t{e}^t}{1\left[e^t.\left[e^t+t{e}^t\right]-e^t.t{e}^t\right]} \\ =\frac{-e^{2t}}{e^{2t}} \\ =-1 \end{gathered}$$

Now integrating this;

 $$\begin{gathered} v_1\left(t\right)=\int -1 dt \\ =-t+C \end{gathered}$$

Step 4: Determine v ' 2 and v 2 .

$$\begin{gathered} v_2\text{'}=\frac{f\left(t\right)y_1\left(t\right)}{a\left[y_1\left(t\right)y{\text{'}}_2\left(t\right)-y{\text{'}}_1\left(t\right)y_2\left(t\right)\right]} \\ =\frac{t^{-1}{e}^t.e^t}{1\left[e^{2t}\right]} \\ =t^{-1} \end{gathered}$$

On integrating, we get:

 $$\begin{gathered} v_2\left(t\right)=\int t^{-1}dt \\ =ln\left|t\right|+C \end{gathered}$$

Therefore, the particular solution is:

$$\begin{gathered} y_p=\left(-t+C\right)e^t+(ln\left|t\right|+C)t{e}^t \\ {y}_p=-t{e}^t+ln\left|t\right|t{e}^t \end{gathered}$$ 

Therefore, the general solution is:

 $$\begin{gathered} y\left(t\right)=y_h\left(t\right)+y_p\left(t\right) \\ y\left(t\right)=c_1{e}^t+c_{2}t{e}^t-t{e}^t+ln\left|t\right|t{e}^t \end{gathered}$$