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

Explanation:

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

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

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: Find a particular solution using variation of parameters

We assume a particular solution of the form $y_p=u_1(x)\cos(x)+u_2(x)\sin(x)$. The Wronskian $W=\begin{vmatrix}\cos(x)&\sin(x)\-\sin(x)&\cos(x)\end{vmatrix}=1$. For the non - homogeneous part $f(x)=\frac{1}{2}-\frac{1}{2}\cos(2x)$, we find $u_1$ and $u_2$ using the formulas: $u_1'=-\frac{\sin(x)f(x)}{W}$ and $u_2'=\frac{\cos(x)f(x)}{W}$. First, for $f(x)=\frac{1}{2}-\frac{1}{2}\cos(2x)$: $u_1'=-\sin(x)(\frac{1}{2}-\frac{1}{2}\cos(2x))=-\frac{1}{2}\sin(x)+\frac{1}{2}\sin(x)\cos(2x)$. Integrating, $u_1=\frac{1}{2}\cos(x)+\frac{1}{6}\cos(3x)-\frac{1}{2}\cos(x)=\frac{1}{6}\cos(3x)$. $u_2'=\cos(x)(\frac{1}{2}-\frac{1}{2}\cos(2x))=\frac{1}{2}\cos(x)-\frac{1}{2}\cos(x)\cos(2x)$. Integrating, $u_2=\frac{1}{2}\sin(x)-\frac{1}{6}\sin(3x)$. So $y_p=\frac{1}{6}\cos(3x)\cos(x)+(\frac{1}{2}\sin(x)-\frac{1}{6}\sin(3x))\sin(x)=\frac{1}{2}-\frac{1}{3}\cos(2x)$.

Step4: General solution

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

Answer:

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