Q6E - 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. $\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{9}\mathit{y}\mathbf{=}\mathit{s}\mathit{e}{\mathit{c}}^{\mathbf{2}}\mathbf{\left(}\mathbf{3}\mathbf{t}\mathbf{\right)}$

Step-by-Step Solution

Answer

The general solution is: 

$y\left(t\right)=c_1\cos 3t+c_2\sin 3t+\left(-\frac{1}{9}\sec 3t\right)\cos 3t+\left(\frac{1}{9}\ln \left|\frac{1+\tan \frac{3t}{2}}{1-\tan \frac{3t}{2}}\right|\right)\sin 3t$

Step 1: Find a particular solution.

The homogenous equation is $r^2+9=0$.

 

Two independent solutions are $r=\pm 3i$.

 

Then $y_1=\cos 3t,y_2=\sin 3t$

$y_h\left(t\right)=c_1\cos 3t+c_2\sin 3t$

 

The particular solution is $y_p=v_1\left(t\right)\cos 3t+v_2\left(t\right)\sin 3t$

Step 2: evaluate v 1 and v 2

Here $y_p=v_1\left(t\right)\cos 3t+v_2\left(t\right)\sin 3t$

 

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 $y_p=0$.

$$\begin{gathered} v_1\text{'}\cos 3t+v_2\text{'}\sin 3t=0 \\ -3v_1\text{'}\sin 3t+3v_2\text{'}\cos 3t=\frac{f}{a} \\ -3v_1\text{'}\sin 3t+3v_2\text{'}\cos 3t={\sec }^{2}3t \end{gathered}$$

Step 3: Find v ' 1 and v 1

$$\begin{gathered} v_1\text{'}=\frac{-f\left(t\right)y_2\left(t\right)}{a\left[y_1\left(t\right)y{\text{'}}_2\left(t\right)-y{\text{'}}_1\left(t\right)y_2\left(t\right)\right]} \\ =\frac{-{\sec }^{2}3t.\sin 3t}{3\left[{\cos }^{2}3t.+{\sin }^{2}3t\right]} \\ =\frac{-1}{3}\tan 3t\sec 3t \end{gathered}$$

Now integrating this.

 $$\begin{gathered} v_1\left(t\right)=\int \frac{-1}{3}\tan 3t\sec 3tdt \\ =-\frac{1}{9}\sec 3t+C \end{gathered}$$

Step 4: Determine v ' 2 and v 2 .

$$\begin{gathered} v_2\text{'}=\frac{f\left(t\right)y_1\left(t\right)}{a\left[y_1\left(t\right)y{\text{'}}_2\left(t\right)-y{\text{'}}_1\left(t\right)y_2\left(t\right)\right]} \\ =\frac{{\sec }^{2}3t.\cos 3t}{3\left[{\cos }^{2}3t.+{\sin }^{2}3t\right]} \\ =\sec 3t \end{gathered}$$

Integrate this.

$$\begin{gathered} v_2\left(t\right)=\int \sec 3tdt \\ =\frac{1}{9}\ln \left|\frac{1+\tan \frac{3t}{2}}{1-\tan \frac{3t}{2}}\right|+C \end{gathered}$$ 

Thus, a particular solution is:

$$\begin{gathered} y_p=\left(-\frac{1}{9}\sec 3t+C\right)\cos 3t+\left(\frac{1}{9}\ln \left|\frac{1+\tan \frac{3t}{2}}{1-\tan \frac{3t}{2}}\right|+C\right)\sin 3t \\ {y}_p=\left(-\frac{1}{9}\sec 3t\right)\cos 3t+\left(\frac{1}{9}\ln \left|\frac{1+\tan \frac{3t}{2}}{1-\tan \frac{3t}{2}}\right|\right)\sin 3t \end{gathered}$$

Therefore, the general solution is: 

 $$\begin{gathered} y\left(t\right)=y_h\left(t\right)+y_p\left(t\right) \\ y\left(t\right)=c_{1}cos3t+c_{2}sin3t+\left(-\frac{1}{9}sec3t\right)cos3t+\left(\frac{1}{9}ln\left|\frac{1+tan\frac{3t}{2}}{1-tan\frac{3t}{2}}\right|\right)sin3t \end{gathered}$$