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Can someone help me to solve the second order differential equation, I'm stuck at finding the particular solution by using undetermined coefficient. $$y''+y'+y=\sin^2 x$$ Is it the particular solution have the form such that $Y(x)=A\sin^2 x + B\cos^2 x$ ?

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  • $\begingroup$ Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. $\endgroup$ – José Carlos Santos Dec 14 '17 at 10:13
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Hint:

Consider $$\sin^2x=1-\cos^2x=1-\frac{\cos(2x)+1}{2}$$

I'll put down the solution in a while if you update your further attempt.

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No, there is no particular solution of the form $A\sin^2 (x) + B\cos^2 (x)$ (see the P.S. below).

Note that $\sin^2(x)=\frac{1}{2}(1-\cos(2x))$. Moreover, $0$ and $2i$, complex numbers associated to the functions $1$ and $\cos(2x)$) are not roots of the characteristic polynomial $z^2+z+1$, then a particular solution should have the form (see LINK): $$A\cdot 1+B\cdot \sin(2x)+C\cdot \cos(2x)$$ where $A,B,C$ are constants to be found. Can you take it from here?

P.S. Note that the dimension of $\text{span}\left(\{1,\sin(2x),\cos(2x)\}\right)$ is $3$, whereas the dimension of $\text{span}\left(\{\sin^2(x),\cos^2(x)\}\right)=\text{span}\left(\{1,\cos(2x)\}\right)$ is $2$. So why is there no particular solution of the form $A\sin^2 (x) + B\cos^2 (x)$?

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