Q17E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problems 11–18, find a general solution to the differential equation.

$\frac{\mathbf{1}}{\mathbf{2}}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{=}\mathbf{tan}\mathbf{2}\mathbf{t}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{e}}^{\mathbf{t}}$

Step-by-Step Solution

Answer

The general solution is $y\left(t\right)=c_1\cos 2t+c_2\sin 2t+-\cos 2t\frac{\ln \left|\sec 2t+\tan 2t\right|}{2}-\frac{e^t}{5}$.

Step 1: Find a particular solution.

The differential equation $\frac{1}{2}y\text{'}\text{'}+2y=\tan 2t-\frac{1}{2}{e}^t$

It can be written as $y\text{'}\text{'}+4y=2\tan 2t-e^t$

 

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

 

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

 

Then $y_1=\cos 2t,v_2=\sin 2t$

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

The particular solution is $y_p=v_1\left(t\right)cos2t+v_2\left(t\right)sin2t$.

Step 2: Evaluate v 1     and     v 2

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

Here $f=\tan 2t-\frac{1}{2}{e}^t$ and $a=1$

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 by derivative then:

$\begin{array}{rcl}{v}_1\text{'}\cos 2t+v_2\text{'}\sin 2t& =& 0\\ -2v_1\text{'}\sin 2t+2v_2\cos 2t& =& \frac{f}{a}\\ -2v_1\text{'}\sin 2t+2v_2\cos 2t& =& 2\tan 2t-e^t\end{array}$ 

Step 3: Find v 1 ' and v 1

$\begin{array}{rcl}{v}_1\text{'}& =& \frac{-f\left(t\right)y_2\left(t\right)}{a\left[y_1\left(t\right)y{\text{'}}_2\left(t\right)-y{\text{'}}_1\left(t\right)y_2\left(t\right)\right]}\\ & =& \frac{-\left(2\tan 2t\right).\sin 2t}{2\left[{\cos }^{2}t+{\sin }^{2}t\right]}\\ & =& -\left(\tan 2t\right).\sin 2t\end{array}$

Now integrating this;

$\begin{array}{rcl}{v}_1\left(t\right)& =& \int -\left(\tan 2t\right).\sin 2 tdt\\ & =& \frac{\ln \left|\sec 2t+\tan 2t\right|}{2}+\frac{\sin 2t}{2}\end{array}$

Step 4: Determine v 2 ' and v 2

$\begin{array}{rcl}{v}_2\text{'}& =& \frac{f\left(t\right)y_1\left(t\right)}{a\left[y_1\left(t\right)y{\text{'}}_2\left(t\right)-y{\text{'}}_1\left(t\right)y_2\left(t\right)\right]}\\ & =& \frac{-\left(2\tan 2t\right).\cos 2t}{2\left[{\cos }^{2}t+{\sin }^{2}t\right]}\\ & =& \left(\tan 2t\right).\cos 2t\\ & =& \sin 2t\end{array}$

Integrate this.

$\begin{array}{rcl}{y}_2\left(t\right)& =& \int \sin 2 tdt\\ & =& -\frac{1}{2}\cos 2t+C\end{array}$ 

Thus, a particular solution is:

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

And the general solution is:

$$\begin{gathered} y\left(t\right)=y_h\left(t\right)+y_p\left(t\right) \\ y\left(t\right)=c_1\cos 2t+c_2\sin 2t+-\cos 2t\frac{\ln \left|\sec 2t+\tan 2t\right|}{2} \end{gathered}$$

Step 5: Find a solution by e t 2

The particular solution is: 

$\begin{array}{rcl}{y}_p& =& A{e}^t\\ y{\text{'}}_p& =& A{e}^t\\ y\text{'}{\text{'}}_p& =& A{e}^t\end{array}$

 

Substitute all values in the differential equation.

$\begin{array}{rcl}\frac{A{e}^t}{2}+2A{e}^t& =& -\frac{1}{2}{e}^t\\ A& =& -\frac{1}{5}\end{array}$ 

Therefore, the final answer is $y\left(t\right)=c_{1}cos2t+c_{2}sin2t+-cos2t\frac{ln\left|sec2t+tan2t\right|}{2}-\frac{e^t}{5}$.