Q8E - 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.

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathit{y}\mathbf{=}\mathit{c}\mathit{s}{\mathit{c}}^{\mathbf{2}}\mathbf{\left(}\mathbf{2}\mathit{t}\mathbf{\right)}$

Step-by-Step Solution

Answer

The general solution is $y\left(t\right)=c_1\cos 2t+c_2\sin 2t+\frac{1}{4}\left(\cos 2t\ln \left[\csc 2t+\cot 2t\right]-1\right)$.

Step 1: Find a particular solution.

The homogenous equation is $r^2+4=0$.

 

Two independent solutions are $r=\pm 2i$.

 

Then $y_1=\cos 2t,y_2=\sin 2t$

$y_h\left(t\right)=c_1\cos 2t+c_2\sin 2t$

 

The particular solution is $y_p=v_1\left(t\right)\cos 2t+v_2\left(t\right)\sin 2t$

Step 2: Evaluate v 1 and v 2

Here $y_p=v_1\left(t\right)\cos 2t+v_2\left(t\right)\sin 2t$

 

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 $y_p=0$

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

Step 3: Find v ' 1 and v 1

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

Now integrating this.

 $$\begin{gathered} v_1\left(t\right)=\int -\frac{1}{2}\csc 2 tdt \\ =-\frac{1}{4}\ln \left[\csc 2t+\cot 2t\right]+C \end{gathered}$$

Step 4: Determine v ' 2 and v 2

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

Integrate this.

$$\begin{gathered} v_2\left(t\right)=\int \frac{1}{2}{\csc }^{2}2t.\cos 2 tdt \\ =-\frac{1}{4}\csc 2t+C \end{gathered}$$ 

 

Thus, a particular solution is:

$$\begin{gathered} y_p=\left(-\frac{1}{4}\ln \left[\csc 2t+\cot 2t\right]+C\right)\cos 2t+\left(-\frac{1}{4}\csc 2t+C\right)\sin 2t \\ {y}_p=\frac{1}{4}\left(\cos 2t\ln \left[\csc 2t+\cot 2t\right]-1\right) \end{gathered}$$ 

Therefore, the general solution is:

$$\begin{gathered} y\left(t\right)=y_h\left(t\right)+y_p\left(t\right) \\ y\left(t\right)=c_{1}cos2t+c_{2}sin2t+\frac{1}{4}\left(cos2tln\left[csc2t+cot2t\right]-1\right) \end{gathered}$$