solve the differential equation by variation of parameters. y + y = sec(θ) tan(θ)

Explanation:

Step1: Find complementary - function

The homogeneous equation is $y'' + y=0$. The characteristic equation is $r^{2}+1 = 0$. Solving $r^{2}+1 = 0$, we get $r=\pm i$. So the complementary - function $y_c = C_1\cos\theta+C_2\sin\theta$.

Step2: Assume particular - solution form

Using the method of variation of parameters, we assume $y_p = u_1(\theta)\cos\theta+u_2(\theta)\sin\theta$. We have the following two equations for $u_1'$ and $u_2'$: $u_1'\cos\theta+u_2'\sin\theta = 0$ and $-u_1'\sin\theta + u_2'\cos\theta=\sec\theta\tan\theta$. From $u_1'\cos\theta+u_2'\sin\theta = 0$, we get $u_1'=-u_2'\frac{\sin\theta}{\cos\theta}$. Substitute into $-u_1'\sin\theta + u_2'\cos\theta=\sec\theta\tan\theta$: $u_2'\left(\sin^{2}\theta/\cos\theta+\cos\theta\right)=\sec\theta\tan\theta$. Since $\sin^{2}\theta/\cos\theta+\cos\theta=\frac{\sin^{2}\theta+\cos^{2}\theta}{\cos\theta}=\sec\theta$, then $u_2'=\tan\theta$. Integrating $u_2'=\tan\theta$ gives $u_2 =-\ln|\cos\theta|$. Since $u_1'=-u_2'\frac{\sin\theta}{\cos\theta}$, then $u_1'=\frac{\sin\theta}{\cos\theta}\cdot\tan\theta=\frac{\sin^{2}\theta}{\cos^{2}\theta}=\sec^{2}\theta - 1$. Integrating $u_1'$ gives $u_1=\tan\theta-\theta$. So $y_p=(\tan\theta - \theta)\cos\theta-\ln|\cos\theta|\sin\theta=\sin\theta-\theta\cos\theta-\sin\theta\ln|\cos\theta|$.

Step3: Find general solution

The general solution of the non - homogeneous equation is $y = y_c + y_p=C_1\cos\theta+C_2\sin\theta+\sin\theta-\theta\cos\theta-\sin\theta\ln|\cos\theta|$.

Answer:

$y = C_1\cos\theta+C_2\sin\theta+\sin\theta-\theta\cos\theta-\sin\theta\ln|\cos\theta|$