Q14E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\left(}\mathbf{\theta }\mathbf{\right)}\mathbf{+}\mathbf{y}\mathbf{\left(}\mathbf{\theta }\mathbf{\right)}\mathbf{=}\mathbf{se}{\mathbf{c}}^{\mathbf{3}}\mathbf{\theta }$.

Step-by-Step Solution

Answer

The general solution is $y\left(t\right)=c_1\cos t+c_2\sin t-\left(\frac{1}{2{\sec }^2\theta }\right)\cos \theta +\left(\tan \theta \right)\sin \theta$.

Step 1: Find a particular solution

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

 

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

 

Then $y_1=\cos \theta ,v_2=\sin \theta$

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

The particular solution is $y_p=y_1\left(\theta \right)cos\theta +y_2\left(\theta \right)sin\theta$.

Step 2: Evaluate v 1     and     v 2

Here $y_p=y_1\left(\theta \right)\cos \theta +y_2\left(\theta \right)\sin \theta$

Here $a=1$ and $f={\sec }^3\theta$

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;

$\begin{array}{rcl}{y}_1\text{'}\cos \theta +y_2\text{'}\sin \theta & =& 0\\ -y_1\text{'}\sin \theta +y_2\cos \theta & =& \frac{f}{a}\\ -y_1\text{'}\sin \theta +y_2\cos \theta & =& {\sec }^3\theta \end{array}$

Step 3: Find v 1 ' and v 1

$\begin{array}{rcl}{v}_1\text{'}& =& \frac{-f\left(\theta \right)v_2\left(\theta \right)}{a\left[v_1\left(\theta \right)v{\text{'}}_2\left(\theta \right)-v{\text{'}}_1\left(\theta \right)v_2\left(\theta \right)\right]}\\ & =& \frac{-\left({\sec }^3\theta \right).\sin \theta }{\left[{\cos }^2\theta +{\sin }^2\theta \right]}\\ & =& -\left({\sec }^3\theta \right).\sin t\end{array}$

Now integrating this;

$\begin{array}{rcl}{v}_1\left(t\right)& =& \int -\left({\sec }^3\theta \right).\sin tdt\\ & =& -\frac{1}{2{\sec }^2\theta }+C\end{array}$ 

Step 4: Determine v 2 ' and v 2

$\begin{array}{rcl}{v}_2\text{'}& =& \frac{-f\left(\theta \right)v_2\left(\theta \right)}{a\left[v_1\left(\theta \right)v{\text{'}}_2\left(\theta \right)-v{\text{'}}_1\left(\theta \right)v_2\left(\theta \right)\right]}\\ & =& \frac{-\left({\sec }^3\theta \right).\cos \theta }{\left[{\cos }^2\theta +{\sin }^2\theta \right]}\\ & =& \left({\sec }^2\theta \right)\end{array}$

Integrate this.

$\begin{array}{rcl}{y}_2\left(t\right)& =& \int {\sec }^2\theta dt\\ & =& \tan \theta +C\end{array}$ 

Thus, the particular solution is:

$$\begin{gathered} y_p=\left(-\frac{1}{2x^2}+C\right)\cos \theta +\left(\tan \theta +C\right)\sin \theta \\ {y}_p=\left(-\frac{1}{2{\sec }^2\theta }\right)\cos \theta +\left(\tan \theta \right)\sin \theta \end{gathered}$$

 

Therefore, the general solution is:

$$\begin{gathered} y\left(t\right)=y_h\left(t\right)+y_p\left(t\right) \\ y\left(t\right)=c_{1}cost+c_{2}sint-(\frac{1}{2se{c}^2\theta })cos\theta +(tan\theta )sin\theta \end{gathered}$$