Question

solve the differential equation by variation of parameters. y + y = sin^2(x)

Explanation:

Step1: Find the 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 $y_c = C_1\cos x + C_2\sin x$.

Step2: Assume a particular - solution form

We assume a particular solution of the form $y_p = u_1(x)\cos x+u_2(x)\sin x$.
We have the following two equations for variation of parameters:
$u_1'\cos x+u_2'\sin x = 0$ and $-u_1'\sin x + u_2'\cos x=\sin^{2}x$.
From $u_1'\cos x+u_2'\sin x = 0$, we get $u_1'=-u_2'\frac{\sin x}{\cos x}$.
Substitute into $-u_1'\sin x + u_2'\cos x=\sin^{2}x$:
$u_2'\frac{\sin^{2}x}{\cos x}+u_2'\cos x=\sin^{2}x$.
$u_2'(\frac{\sin^{2}x+\cos^{2}x}{\cos x})=\sin^{2}x$.
Since $\sin^{2}x+\cos^{2}x = 1$, then $u_2'=\sin^{2}x\cos x$.
Integrating $u_2'$ with respect to $x$: $u_2=\int\sin^{2}x\cos xdx$. Let $t = \sin x$, $dt=\cos xdx$, so $u_2=\frac{1}{3}\sin^{3}x$.
From $u_1'=-u_2'\frac{\sin x}{\cos x}$, $u_1'=-\sin^{3}x$. Integrating $u_1'$: $u_1=\cos x-\frac{1}{3}\cos^{3}x$.
So $y_p = (\cos x-\frac{1}{3}\cos^{3}x)\cos x+\frac{1}{3}\sin^{3}x\sin x=\cos^{2}x-\frac{1}{3}\cos^{4}x+\frac{1}{3}\sin^{4}x$.
Using $\sin^{2}x = 1 - \cos^{2}x$, $y_p=\cos^{2}x-\frac{1}{3}\cos^{4}x+\frac{1}{3}(1 - \cos^{2}x)^{2}=\cos^{2}x-\frac{1}{3}\cos^{4}x+\frac{1}{3}(1 - 2\cos^{2}x+\cos^{4}x)=\frac{1}{3}-\frac{1}{3}\cos^{2}x$.

Step3: General solution

The general solution of the non - homogeneous equation $y'' + y=\sin^{2}x$ is $y = y_c + y_p=C_1\cos x + C_2\sin x+\frac{1}{3}-\frac{1}{3}\cos^{2}x$.

Answer:

$y = C_1\cos x + C_2\sin x+\frac{1}{3}-\frac{1}{3}\cos^{2}x$