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

Explanation:

Step1: Rewrite the right - hand side

Use the double - angle formula $\sin^{2}x=\frac{1 - \cos(2x)}{2}$. 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 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 $-u_1'\sin x+u_2'\cos x=\frac{1 - \cos(2x)}{2}$. From the first equation $u_2'=-u_1'\frac{\cos x}{\sin x}$. Substitute into the second equation: $-u_1'\sin x - u_1'\frac{\cos^{2}x}{\sin x}=\frac{1 - \cos(2x)}{2}$. $-u_1'\frac{\sin^{2}x+\cos^{2}x}{\sin x}=\frac{1 - \cos(2x)}{2}$, so $u_1'=-\frac{\sin x(1 - \cos(2x))}{2}$. Integrating $u_1'$: $u_1=\frac{1}{2}\int(-\sin x+\sin x\cos(2x))dx$. We know that $\int\sin x\cos(2x)dx=\frac{1}{2}\int(\sin(3x)-\sin x)dx$. $u_1=\frac{1}{2}(\cos x+\frac{1}{6}\cos(3x)-\frac{1}{2}\cos x)=\frac{1}{4}\cos x+\frac{1}{12}\cos(3x)$. From $u_1'\cos x+u_2'\sin x = 0$, we find $u_2'=\frac{\cos x(1 - \cos(2x))}{2\sin x}$. Integrating $u_2'$ gives $u_2=-\frac{1}{2}\int(\cot x-\cot x\cos(2x))dx$. $u_2 = \frac{1}{2}\ln|\sin x|-\frac{1}{12}\sin(3x)$. The particular solution $y_p=\left(\frac{1}{4}\cos x+\frac{1}{12}\cos(3x)\right)\cos x+\left(\frac{1}{2}\ln|\sin x|-\frac{1}{12}\sin(3x)\right)\sin x$. $y_p=\frac{1}{4}\cos^{2}x+\frac{1}{12}\cos(3x)\cos x+\frac{1}{2}\sin x\ln|\sin x|-\frac{1}{12}\sin(3x)\sin x$. Using trigonometric identities $\cos^{2}x=\frac{1 + \cos(2x)}{2}$, $\cos A\cos B=\frac{1}{2}[\cos(A + B)+\cos(A - B)]$ and $\sin A\sin B=\frac{1}{2}[\cos(A - B)-\cos(A + B)]$. $y_p=\frac{1}{8}+\frac{1}{8}\cos(2x)+\frac{1}{24}\cos(4x)+\frac{1}{24}\cos(2x)+\frac{1}{2}\sin x\ln|\sin x|-\frac{1}{24}\cos(2x)+\frac{1}{24}\cos(4x)$. $y_p=\frac{1}{8}+\frac{1}{8}\cos(2x)+\frac{1}{12}\cos(4x)+\frac{1}{2}\sin x\ln|\sin x|$. The general solution is $y = y_c + y_p=C_1\cos x+C_2\sin x+\frac{1}{8}+\frac{1}{8}\cos(2x)+\frac{1}{12}\cos(4x)+\frac{1}{2}\sin x\ln|\sin x|$.

Answer:

$y = C_1\cos x+C_2\sin x+\frac{1}{8}+\frac{1}{8}\cos(2x)+\frac{1}{12}\cos(4x)+\frac{1}{2}\sin x\ln|\sin x|$