Q20E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Use the method of variation of parameters to show that $\mathbf{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}\mathbf{cost}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\mathbf{sint}\mathbf{+}\underset{\mathbf{0}}{\overset{\mathbf{t}}{\int }}\mathbf{f}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{sin}\mathbf{(}\mathbf{t}\mathbf{-}\mathbf{s}\mathbf{)}\mathbf{ds}$ is a general solution to the differential

equation $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$, where f(t) is a continuous function on $\mathbf{(}\mathbf{-}\mathbf{\infty }\mathbf{,}\mathbf{\infty }\mathbf{)}$.[Hint: Use the trigonometric identity $\mathbf{sin}\mathbf{(}\mathbf{t}\mathbf{-}\mathbf{s}\mathbf{)}\mathbf{=}\mathbf{sintcoss}\mathbf{-}\mathbf{sinscost}$.]

Step-by-Step Solution

Answer

The general solution is $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_1\mathrm{cost}+{\mathrm{c}}_2\mathrm{sint}-\ln \left|\mathrm{sect}+\mathrm{tant}\right|.\mathrm{cost}+\frac{1}{10}{\mathrm{e}}^{3\mathrm{t}}-1$.

 

Step 1: Find particular solution

The homogenous equation is ${\mathrm{r}}^2+1=0$.

 

Two independent solutions are $\mathrm{r}=\pm \mathrm{i}$.

 

Then ${\mathrm{y}}_1=\mathrm{cost},{\mathrm{y}}_2=\mathrm{sint}$

${\mathrm{y}}_{\mathrm{h}}\left(\mathrm{t}\right)={\mathrm{c}}_1\mathrm{cost}+{\mathrm{c}}_2\mathrm{sint}$ 

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

Step 2: Evaluate v 1   and   v 2

Here ${\mathrm{y}}_{\mathrm{p}}={\mathrm{v}}_1\left(\mathrm{t}\right)\mathrm{cost}+{\mathrm{v}}_2\left(\mathrm{t}\right)\mathrm{sint}$

 

And referring to (9) and solve the system then

$$\begin{gathered} {\mathrm{v}}_1\text{'}\mathrm{cost}+{\mathrm{v}}_2\text{'}\mathrm{sint}=0 \\ -{\mathrm{v}}_1\text{'}\mathrm{sint}+{\mathrm{v}}_2\mathrm{cost}=\frac{\mathrm{f}}{\mathrm{a}} \\ -{\mathrm{v}}_1\text{'}\mathrm{sint}+{\mathrm{v}}_2\mathrm{cost}=\mathrm{f}\left(\mathrm{t}\right) \end{gathered}$$ 

Step 3: Find v 1 ' and v 1

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

Now integrating this.

${\mathrm{v}}_1\left(\mathrm{t}\right)=\int -\mathrm{f}\left(\mathrm{t}\right).\mathrm{sintdt}$

 

Step 4: Determine v 2 ' and v 2

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

Integrate this.

${\mathrm{v}}_2\left(\mathrm{t}\right)=\int \mathrm{f}\left(\mathrm{t}\right)\mathrm{costdt}$ 

Thus the particular solution is:

${\mathrm{y}}_{\mathrm{p}}=\left(\int -\mathrm{f}\left(\mathrm{t}\right).\mathrm{sintdt}\right)\mathrm{cost}+\left(\int \mathrm{f}\left(\mathrm{t}\right)\mathrm{costdt}\right)\mathrm{sint}$

 

And the general solution is: 

$$\begin{gathered} \mathrm{y}\left(\mathrm{t}\right)={\mathrm{y}}_{\mathrm{h}}\left(\mathrm{t}\right)+{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right) \\ \mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_1\mathrm{cost}+{\mathrm{c}}_2\mathrm{sint}+\left(\int -\mathrm{f}\left(\mathrm{t}\right).\mathrm{sintdt}\right)\mathrm{cost}+\left(\int \mathrm{f}\left(\mathrm{t}\right)\mathrm{costdt}\right)\mathrm{sint} \\ \mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_1\mathrm{cost}+{\mathrm{c}}_2\mathrm{sint}+\underset{0}{\overset{\mathrm{s}}{\int }}\mathrm{f}\left(\mathrm{s}\right)\sin (\mathrm{t}-\mathrm{s})\mathrm{ds} \end{gathered}$$

This is the required result.