Q4E - 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.

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

Step-by-Step Solution

Answer

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

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 $y_p=0$.

$$\begin{gathered} v_1\text{'}{e}^{-t}+v_2\text{'}t{e}^{-t}=0 \\ {v}_1\text{'}{e}^{-t}+v_2\text{'}\left[e^{-t}-t{e}^{-t}\right]=\frac{f}{a} \\ {v}_1\text{'}{e}^{-t}+v_2\text{'}\left[e^{-t}-t{e}^{-t}\right]=e^{-t} \\ v_1\text{'}+v_2\text{'}\left[1-t\right]=1 \end{gathered}$$

Step 3: Find v ' 1 and v 1

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

Now integrating this;

$$\begin{gathered} v_1\left(t\right)=\int -t dt \\ =-\frac{t^2}{2}+C \end{gathered}$$

Step 4: Determine v ' 2 and v 2

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

Integrate this:

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

 

Therefore, the particular solution is:

$$\begin{gathered} y_p=\left(-\frac{t^2}{2}+C\right)e^{-t}+\left(t+C\right)t{e}^{-t} \\ {y}_p=-\frac{t^2}{2}{e}^{-t} \end{gathered}$$ 

Hence, the general solution is:

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