Q6E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problems 1-6, use the method of variation of parameters to determine a particular solution to the given equation. ${\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{=}\mathit{s}\mathit{e}\mathit{c}\mathit{\theta }\mathit{t}\mathit{a}\mathit{n}\mathit{\theta }\mathbf{,}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{0}\mathbf{<}\mathit{\theta }\mathbf{<}\mathit{\pi }\mathbf{/}\mathbf{2}$

Step-by-Step Solution

Answer

The particular solution is $y_p=\sec \theta -\cos \theta \ln \left|\sec \theta \right|-\sin \theta \tan \theta +\theta \sin \theta$

Step 1: Definition

Variation of parameters, also known as variation of constants, is a general method to solve inhomogeneous linear ordinary differential equations.

Step 2: Find complementary solution

The given equation is:$y^{\text{'}\text{'}\text{'}}+y^{\text{'}}=\sec \theta \tan \theta , 0<\theta <\pi /2$

The auxiliary equation is $m^3+m=0$

$$\begin{gathered} m\left(m^2+1\right)=0 \\ m=0\text{ or }{m}^2+1=0 \\ m=0\text{ or }m=\pm i \end{gathered}$$

The complimentary solution is, $y_c=c_1+c_2\cos \theta +c_3\sin \theta$

The fundamental solution set is $\{1,\cos \theta ,\sin \theta \}$

Step 3: Calculate Wornkians

$$\begin{gathered} W\left[\begin{array}{ccc}1& \cos \theta & \sin \theta \end{array}\right]=\left|\begin{array}{ccc}1& \cos \theta & \sin \theta \\ 0& -\sin \theta & \cos \theta \\ 0& -\cos \theta & -\sin \theta \end{array}\right| \\ =1 \\ {W}_1\left[\theta \right]=W_1(-1{)}^{3-1}W\left[\begin{array}{cc}\cos \theta & \sin \theta \end{array}\right] \\ =1 \\ {W}_2\left[\theta \right]=(-1{)}^{3-2}W\left[\begin{array}{cc}1& \sin \theta \end{array}\right] \\ =-\left|\begin{array}{cc}1& \sin \theta \\ 0& cos\theta \end{array}\right| \\ {W}_3\left[\theta \right]=(-1{)}^{3-3}W\left[\begin{array}{cc}1& cos\theta \end{array}\right] \\ =-\sin \theta \end{gathered}$$

Step 4: For particular solution

The particular solution is of the form $y_p\left(x\right)=v_1\left(x\right)\left(1\right)+v_2\left(x\right)\cos \theta +v_3\left(x\right)\left(\sin \theta \right)$

The particular solution is given by: 

$$\begin{gathered} y_p=1\int \frac{\sec \theta \tan \theta }{1}d\theta +\cos \theta \int \frac{-\cos \theta }{1}\sec \theta \tan \theta d\theta +\sin \theta \int \frac{-\sin \theta }{1}\left(\sec \theta \tan \theta \right)d\theta \\ =\int \sec \theta \tan \theta d\theta -\cos \theta \int \tan \theta d\theta -\sin \theta \int {\tan }^2\theta d\theta \\ =sec\theta -\cos \theta \ln \left|\sec \theta \right|-\sin \theta (\tan \theta -\theta ) \\ =\sec \theta -\cos \theta \ln \left|\sec \theta \right|-\sin \theta \tan \theta +\theta \sin \theta \end{gathered}$$