Question

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

Explanation:

Step1: Rewrite $\sin^{2}(x)$

Use the double - angle formula $\sin^{2}(x)=\frac{1 - \cos(2x)}{2}$. So the differential equation becomes $y''+y=\frac{1}{2}-\frac{\cos(2x)}{2}$.

Step2: Find the complementary function

The homogeneous equation is $y'' + y = 0$. The characteristic equation is $r^{2}+1 = 0$, which gives $r=\pm i$. So the complementary function $y_c = C_1\cos(x)+C_2\sin(x)$.

Step3: Use variation of parameters

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 $u_1'$ and $u_2'$:
$u_1'\cos(x)+u_2'\sin(x)=0$ and $-u_1'\sin(x)+u_2'\cos(x)=\frac{1}{2}-\frac{\cos(2x)}{2}$
Solving this system for $u_1'$ and $u_2'$:
From the first equation $u_2'=-\frac{u_1'\cos(x)}{\sin(x)}$. Substitute into the second equation:
$-u_1'\sin(x)-\frac{u_1'\cos^{2}(x)}{\sin(x)}=\frac{1}{2}-\frac{\cos(2x)}{2}$
$u_1'=-\frac{\sin(x)}{2}+\frac{\sin(x)\cos(2x)}{2}$
Integrating $u_1'$:
$u_1=\frac{\cos(x)}{2}-\frac{1}{6}\cos(3x)$
And $u_2'=\frac{\cos(x)}{2}-\frac{\cos(x)\cos(2x)}{2}$
Integrating $u_2'$:
$u_2=\frac{\sin(x)}{2}-\frac{1}{6}\sin(3x)$
So $y_p=\frac{1}{2}-\frac{1}{6}\cos(2x)$

Step4: General solution

The general solution of the non - homogeneous equation is $y = y_c + y_p=C_1\cos(x)+C_2\sin(x)+\frac{1}{2}-\frac{1}{6}\cos(2x)$

Answer:

$y = C_1\cos(x)+C_2\sin(x)+\frac{1}{2}-\frac{1}{6}\cos(2x)$