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

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 based on the method of variation of parameters:
$u_1'\cos(x)+u_2'\sin(x)=0$ and $-u_1'\sin(x)+u_2'\cos(x)=\frac{1 - \cos(2x)}{2}$.
From the first equation, $u_2'=-u_1'\frac{\cos(x)}{\sin(x)}$. Substitute it into the second equation:
$-u_1'\sin(x)-u_1'\frac{\cos^{2}(x)}{\sin(x)}=\frac{1 - \cos(2x)}{2}$.
$-u_1'\frac{\sin^{2}(x)+\cos^{2}(x)}{\sin(x)}=\frac{1 - \cos(2x)}{2}$.
Since $\sin^{2}(x)+\cos^{2}(x) = 1$, we have $u_1'=-\frac{\sin(x)(1 - \cos(2x))}{2}$.
Integrate $u_1'$:

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LATEXBLOCK0
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From $u_1'\cos(x)+u_2'\sin(x)=0$, we find $u_2$.
$u_2'=\frac{\cos(x)(1 - \cos(2x))}{2\sin(x)}$.
Integrate $u_2'$:

$$
LATEXBLOCK1
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The particular solution $y_p = u_1(x)\cos(x)+u_2(x)\sin(x)$.

$$
LATEXBLOCK2
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The general solution $y = y_c + y_p=C_1\cos(x)+C_2\sin(x)+\frac{\cos^{4}(x)}{3}-\cos^{2}(x)+\frac{\sin^{3}(x)}{2}$.

Answer:

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