Q9E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problems 9 and 10, find a particular solution first by undetermined coefficients, and then by variation of parameters. Which method was quicker?

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathit{y}\mathbf{=}\mathbf{2}\mathit{t}\mathbf{+}\mathbf{4}$

Step-by-Step Solution

Answer

In both cases, the particular solution is the same $y_p=-2t-4$. But the method's undetermined coefficient was a bit quicker.

Step 1: Find a particular solution by variation of parameter.

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

 

Two independent solutions are $r=\pm 1$.

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 , and v 2 , v ' 1 and v1 , v ' 2 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 $y_p=0$

$$\begin{gathered} v_1\text{'}{e}^t+v_2\text{'}t{e}^t=0 \\ {v}_1\text{'}{e}^t+v_2\text{'}\left[e^t+t{e}^t\right]=\frac{f}{a} \\ {v}_1\text{'}{e}^t+v_2\text{'}\left[e^t+t{e}^t\right]=2t+4 \end{gathered}$$

$$\begin{gathered} v_1\text{'}=\frac{-f\left(t\right)y_2\left(t\right)}{a\left[y_1\left(t\right)y{\text{'}}_2\left(t\right)-y{\text{'}}_1\left(t\right)y_2\left(t\right)\right]} \\ =\frac{-\left(2t+4\right).t{e}^t}{\left[e^t\left[e^{-t}-t{e}^{-t}\right]-e^t.t{e}^t\right]} \\ =\frac{t+2}{e^t} \end{gathered}$$

Now integrating this.

$$\begin{gathered} v_1\left(t\right)=\int \frac{t+3}{e^t} dt \\ =-e^{-t}(t+3)+C \end{gathered}$$

$$\begin{gathered} v_2\text{'}=\frac{f\left(t\right)y_1\left(t\right)}{a\left[y_1\left(t\right)y{\text{'}}_2\left(t\right)-y{\text{'}}_1\left(t\right)y_2\left(t\right)\right]} \\ =\frac{\left(2t+4\right).e^t}{\left[e^t\left[e^{-t}-t{e}^{-t}\right]-e^t.t{e}^t\right]} \\ =-\left(t+2\right)e^t \end{gathered}$$

Integrate this.

$$\begin{gathered} v_2\left(t\right)=\int -\left(t+2\right)e^{t}dt \\ =-e^t\left(t+1\right)+C \end{gathered}$$ 

Thus, a particular solution is when $C=0$

 $$\begin{gathered} y_p=-e^{-t}(t+3)+C-e^t(t+1)+C \\ {y}_p=-2t-4 \end{gathered}$$

And the general solution is:

$$\begin{gathered} y\left(t\right)=y_h\left(t\right)+y_p\left(t\right) \\ y\left(t\right)=c_1{e}^t+c_{2}t{e}^{-t}-2t-4 \end{gathered}$$

Step 3: Determine a particular solution by an undetermined coefficient.

This method gives a solution to the differential equations in the form,

 $y_p\left(t\right)=t^s\left(A_m{t}^m+......+A_{1}t+A_o\right)e^{rt}$

Split the given equation into two parts;

$$\begin{gathered} y\text{'}\text{'}-y=2t \\ y\text{'}\text{'}-y=4 \end{gathered}$$ 

• $S=0$ if r is not a root of the corresponding auxiliary equation.
• $S=1$ if r is the single root of the corresponding auxiliary equation.
• $S=2$ if r is the double root of the corresponding auxiliary equation.

For the first equation $C=2,m=1$, and $r=0$ since if r is not a root of the corresponding auxiliary equation.

 

The form of the particular solution for the equation $y_{p_1}\left(t\right)=A_{1}t+A_o$

$y{\text{'}}_{p_1}\left(t\right)=A_1 \text{and} y\text{'}{\text{'}}_{p_1}\left(t\right)=0$

So, $y\text{'}{\text{'}}_{p_1}-y_{p_1}=0-2A_{1}t-A_o=2t$

By solving $A_1=-2,A_o=0$

The first particular solution is $y_{p_1}\left(t\right)=-2t$

For the second equation $C=4,m=0,r=0$. Again, r is not a root of the corresponding auxiliary equation. 

 

The form of the particular solution for the equation $y_{p_2}\left(t\right)=B$.

$y{\text{'}}_{p_2}\left(t\right)=0 \text{and} y\text{'}{\text{'}}_{p_2}\left(t\right)=0$

 

By solving this, $B=-4$

 

The second particular solution is $y_{p_2}\left(t\right)=-4$.

Therefore, a particular solution is $y_p=-2t-4$.