Question

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

Explanation:

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

We know that $\sin^{2}(x)=\frac{1 - \cos(2x)}{2}$. So the differential equation becomes $y''+y=\frac{1 - \cos(2x)}{2}$.

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 the method of variation of parameters

Let $y_p = u_1(x)\cos(x)+u_2(x)\sin(x)$.
We have the following two equations for $u_1'(x)$ and $u_2'(x)$:
$u_1'\cos(x)+u_2'\sin(x)=0$ and $-u_1'\sin(x)+u_2'\cos(x)=\frac{1 - \cos(2x)}{2}$.
From the first equation $u_2'=-\frac{u_1'\cos(x)}{\sin(x)}$. Substitute it into the second equation:
$-u_1'\sin(x)-\frac{u_1'\cos^{2}(x)}{\sin(x)}=\frac{1 - \cos(2x)}{2}$.
$-u_1'\frac{\sin^{2}(x)+\cos^{2}(x)}{\sin(x)}=\frac{1 - \cos(2x)}{2}$.
Since $\sin^{2}(x)+\cos^{2}(x) = 1$, we have $u_1'=-\frac{\sin(x)(1 - \cos(2x))}{2}$.
Integrate $u_1'$:
$u_1=\frac{1}{2}\int(-\sin(x)+\sin(x)\cos(2x))dx$.
We know that $\sin(A)\cos(B)=\frac{\sin(A + B)+\sin(A - B)}{2}$, so $\sin(x)\cos(2x)=\frac{\sin(3x)-\sin(x)}{2}$.
$u_1=\frac{1}{2}\int(-\sin(x)+\frac{\sin(3x)-\sin(x)}{2})dx=\frac{1}{2}\int(-\frac{3}{2}\sin(x)+\frac{\sin(3x)}{2})dx=\frac{3}{4}\cos(x)-\frac{1}{12}\cos(3x)$.
From $u_1'\cos(x)+u_2'\sin(x)=0$, we get $u_2'=\frac{\cos(x)(1 - \cos(2x))}{2\sin(x)}$.
Integrate $u_2'$:
$u_2=\frac{1}{2}\int\frac{\cos(x)(1 - \cos(2x))}{\sin(x)}dx$.
Let $t=\sin(x)$, then $dt=\cos(x)dx$.
$u_2=\frac{1}{2}\int\frac{1-(1 - 2t^{2})}{t}dt=\frac{1}{2}\int 2tdt=\frac{1}{2}t^{2}=\frac{1}{2}\sin^{2}(x)$.
The particular solution $y_p = u_1\cos(x)+u_2\sin(x)=\frac{3}{4}\cos^{2}(x)-\frac{1}{12}\cos(3x)\cos(x)+\frac{1}{2}\sin^{3}(x)$.
The general solution $y = y_c + y_p=C_1\cos(x)+C_2\sin(x)+\frac{1}{2}-\frac{1}{4}\cos(2x)-\frac{1}{12}\cos(3x)\cos(x)+\frac{1}{2}\sin^{3}(x)$.

Answer:

$y = C_1\cos(x)+C_2\sin(x)+\frac{1}{2}-\frac{1}{4}\cos(2x)-\frac{1}{12}\cos(3x)\cos(x)+\frac{1}{2}\sin^{3}(x)$