How could I solve the equation $y''+y= \frac{1}{\sin(t)}$? How could I solve the differential equation $y''+y= \frac{1}{\sin(t)}$? 
A short moment, I thought I could use the equation $y''+y= \sin(t)$. I think I have to find the general solution (trivial), but what about the particular solution?
Any idea?
 A: Any solution to a differential equation of the kind $$y''+py'+qy=r$$ can be written as $$y=c_1y_1+c_2y_2+\overline{y}$$ where $y_1$ and $y_2$ are two linearly independent solutions of the homogenous equation $$y''+py'+qy=0$$ and $\overline{y}$ is a particular solution of the original differential equation.
Now here, your homogenous equation is $$y''+y=0$$ Clearly $\cos t$ and $\sin t$ are two solutions. Both are linearly independent. Now to find the partcular solution, use the Method of Variation of parameters. Note that the wronskian($W$) of $\sin t$ and $\cos t$ is $-1$.
Thus, $\overline{y}=u_1y_1+u_2y_2$ where $$u_1=-\int \dfrac {y_2r}{W}\mathrm dt=-\int \dfrac {\cos t\cdot \dfrac {1}{\sin t}}{-1} \mathrm dt=\ln \sin t$$ and $$u_2=\int \dfrac {y_1r}{W}\mathrm dt=\int \dfrac {\sin t\cdot \dfrac{1}{\sin t}}{-1}\mathrm dt=-t$$
Therefore, $\overline{y}=\ln\sin t \cdot \sin t+(-t)\cdot \cos t$.
Thus the general solution is $$y=c_1\sin t+c_2\cos t+\ln\sin t \times \sin t+(-t)\times \cos t$$.
