Q12E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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

$\mathbf{12}\mathbf{.}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{=}{\mathbf{tan}}^{\mathbf{2}}\mathbf{t}$

Step-by-Step Solution

Answer

The general solution is $c_1\cos t+c_2\sin t-\sin t\ln \left|\sec t+\tan t\right|-2$

Step 1: Find a particular solution .

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

 

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

 

Then $y_1=\cos t,y_2=\sin t$

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

The particular solution is $y_p=v_1\left(t\right)cost+v_2\left(t\right)sint$

Step 2: Find v 1 ' and v 1

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

Now integrating this,

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

Step 3: Determine v 2 ' and v 2

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

Integrate this.

$\begin{array}{rcl}{v}_2\left(t\right)& =& \int {\tan }^{2}t.\cos tdt\\ & =& \ln \left|\sec t+\tan t\right|-\sin t+C\end{array}$ 

Thus, a particular solution is:

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

 

Therefore, the general solution is: 

$$\begin{gathered} y\left(t\right)=y_h\left(t\right)+y_p\left(t\right) \\ y\left(t\right)=c_{1}cost+\pm c_{2}sint-sintln\left|sect+tant\right|-2 \end{gathered}$$