Answer to Question #245173 in Differential Equations for lilly

Question

y''+y=sin^2(x)

Answer 1

$y^{\prime\prime}+y=\sin^{2}(x)$ (1)

The general solution will be the sum of the complementary solution and particular solution.

Find the complementary solution for the homogeneous equation:

$y^{\prime\prime}+y=0$.

Compose the characteristic equation:

$\lambda^{2}+1=0\implies \lambda=i$ or $\lambda=-i$.

For complementary solution:

$y_{c}=y_{1}(x)+y_{2}(x)$, where $y_{1}(x)=C_{1}\cos(x)$ and $y_{2}(x)=C_{2}\sin(x).$

$y_{c}=C_{1}\cos(x)+C_{2}\sin(x)$

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Find the particular solution for (1).

$\sin^{2}(x)=\frac{1}{2}-\frac{1}{2}\cos(2x)\implies y^{\prime\prime}+y=\frac{1}{2}-\frac{1}{2}\cos(2x)$.

It's clear that $y_{p}=\frac{1}{2}+\alpha\cos(2x)$, $\alpha - constant$.

Find $\alpha.$

$y_p^{\prime\prime}+y_p=4\alpha(-\cos(2x))+\frac{1}{2}+\alpha\cos(2x)=\frac{1}{2}-\frac{1}{2}\cos(2x)$.

$\boxed{\alpha=\frac{1}{6}}$

$y_{p}=\frac{1}{2}+\frac{1}{6}\cos(2x)$.

The general solution is:

$y(x)=y_{c}(x)+y_{p}(x)=C_{1}\cos(x)+C_{2}\sin(x)+\frac{1}{2}+\frac{1}{6}\cos(2x)$.

$\boxed{y=C_{1}\cos(x)+C_{2}\sin(x)+\frac{1}{2}+\frac{1}{6}\cos(2x)}$