\[ -\sin (n x)+y''(x)+y(x)=0 \] ✓ Mathematica : cpu = 0.0959813 (sec), leaf count = 45
\[\left \{\left \{y(x)\to \frac {\cos ^2(x) (-\sin (n x))-\sin ^2(x) \sin (n x)}{n^2-1}+c_1 \cos (x)+c_2 \sin (x)\right \}\right \}\] ✓ Maple : cpu = 0.064 (sec), leaf count = 26
\[y \relax (x ) = \sin \relax (x ) c_{2}+\cos \relax (x ) c_{1}-\frac {\sin \left (n x \right )}{n^{2}-1}\]
Hand solution
\begin {equation} y^{\prime \prime }+y=\sin nx\tag {1} \end {equation} We start by solving the homogeneous equation \[ y^{\prime \prime }+y=0 \] Let \(y=e^{\lambda x}\), substitution in above gives\begin {align*} \lambda ^{2}e^{\lambda x}+e^{\lambda x} & =0\\ \lambda ^{2}+1 & =0 \end {align*}
Hence \(\lambda =\pm i\), therefore the solution is\begin {align*} y_{h} & =Ae^{ix}+Be^{-ix}\\ & =A\left (\cos x+i\sin x\right ) +B\left (\cos x-i\sin x\right ) \\ & =\cos x\left (A+B\right ) +\sin x\left (Ai-iB\right ) \\ & =\cos x\left (A+B\right ) +\sin x\left (i\left (A-B\right ) \right ) \end {align*}
Let \(A+B=c_{1},i\left (A-B\right ) =c_{2}\) hence\[ y_{h}=c_{1}\cos x+c_{2}\sin x \] Now we solve for the particular solution using variation of parameters. Let \begin {align*} y_{p} & =u_{1}\relax (x) \cos x+u_{2}\relax (x) \sin x\\ y_{p}^{\prime } & =u_{1}^{\prime }\cos x-u_{1}\sin x+u_{2}^{\prime }\sin x+u_{2}\cos x\\ & =u_{2}\cos x-u_{1}\sin x+u_{1}^{\prime }\cos x+u_{2}^{\prime }\sin x \end {align*}
Let first condition be \begin {equation} u_{1}^{\prime }\cos x+u_{2}^{\prime }\sin x=0\tag {2} \end {equation}
Hence\begin {align*} y_{p}^{\prime } & =u_{2}\cos x-u_{1}\sin x\\ y_{p}^{\prime \prime } & =u_{2}^{\prime }\cos x-u_{2}\sin x-u_{1}^{\prime }\sin x-u_{1}\cos x \end {align*}
Substituting in (1) gives \begin {align} y_{p}^{\prime \prime }+y_{p} & =\sin nx\nonumber \\ u_{2}^{\prime }\cos x-u_{2}\sin x-u_{1}^{\prime }\sin x-u_{1}\cos x+u_{1}\cos x+u_{2}\sin x & =\sin nx\nonumber \\ u_{2}^{\prime }\cos x-u_{1}^{\prime }\sin x & =\sin nx\tag {3} \end {align}
So we have two equations (1)(2) to solve for \(u_{1},u_{2}\)\begin {align*} u_{1}^{\prime }\cos x+u_{2}^{\prime }\sin x & =0\\ u_{2}^{\prime }\cos x-u_{1}^{\prime }\sin x & =\sin nx \end {align*}
From the first equation\begin {equation} u_{1}^{\prime }=-u_{2}^{\prime }\frac {\sin x}{\cos x}\tag {4} \end {equation} Substituting in the second equation\begin {align*} u_{2}^{\prime }\cos x-\left (-u_{2}^{\prime }\frac {\sin x}{\cos x}\right ) \sin x & =\sin nx\\ u_{2}^{\prime }\left (\cos x+\frac {\sin x}{\cos x}\sin x\right ) & =\sin nx\\ u_{2}^{\prime }\left (\frac {\cos ^{2}x+\sin ^{2}x}{\cos x}\right ) & =\sin nx\\ u_{2}^{\prime } & =\cos x\sin nx \end {align*}
Hence \begin {align*} u_{2} & =\int \cos x\sin \left (nx\right ) dx\\ & =\frac {-n\cos x\cos \left (nx\right ) -\sin x\sin \left (nx\right ) }{n^{2}-1} \end {align*}
From (4)\begin {align*} u_{1}^{\prime } & =-\cos x\sin nx\frac {\sin x}{\cos x}\\ u_{1} & =-\int \sin \left (nx\right ) \sin xdx\\ & =\frac {n\cos \left (nx\right ) \sin x-\cos x\sin \left (nx\right ) }{n^{2}-1} \end {align*}
Since \(y_{p}=u_{1}\relax (x) \cos x+u_{2}\relax (x) \sin x\) then\begin {align*} y_{p} & =\left (\frac {n\cos \left (nx\right ) \sin x-\cos x\sin \left ( nx\right ) }{n^{2}-1}\right ) \cos x+\left (\frac {-n\cos x\cos \left ( nx\right ) -\sin x\sin \left (nx\right ) }{n^{2}-1}\right ) \sin x\\ & =\frac {n\cos \left (nx\right ) \cos x\sin x-\cos ^{2}x\sin \left (nx\right ) -n\cos x\sin x\cos \left (nx\right ) -\sin ^{2}x\sin \left (nx\right ) }{n^{2}-1}\\ & =\frac {-\sin \left (nx\right ) \left (\cos ^{2}x+\sin ^{2}x\right ) }{n^{2}-1}\\ & =\frac {\sin \left (nx\right ) }{1-n^{2}} \end {align*}
Therefore, the full solution is (for \(n^{2}\neq 1\))\begin {align*} y & =y_{h}+y_{p}\\ & =c_{1}\cos x+c_{2}\sin x+\frac {\sin \left (nx\right ) }{1-n^{2}} \end {align*}
Solution using undetermined coefficients: Since RHS is \(\sin nx\) we guess \(y_{p}=A\cos \left (nx\right ) +B\sin \left (nx\right ) \), therefore
\begin {align*} y_{p}^{\prime } & =-An\sin \left (nx\right ) +Bn\cos \left (nx\right ) \\ y_{p}^{\prime \prime } & =-An^{2}\cos \left (nx\right ) -Bn^{2}\sin \left ( nx\right ) \end {align*}
Plug into the ODE gives
\begin {align*} y_{p}^{\prime \prime }+y_{p} & =\sin nx\\ -An^{2}\cos \left (nx\right ) -Bn^{2}\sin \left (nx\right ) +A\cos \left ( nx\right ) +B\sin \left (nx\right ) & =\sin nx\\ \cos \left (nx\right ) \left (-An^{2}+A\right ) +\sin \left (nx\right ) \left (-Bn^{2}+B\right ) & =\sin \left (nx\right ) \end {align*}
Hence \(-Bn^{2}-B=1\) and \(-An^{2}+A=0\). Therefore \(A=0\) and from the first equation
\begin {align*} B\left (n^{2}+1\right ) & =-1\\ B & =\frac {-1}{n^{2}+1} \end {align*}
Hence \begin {align*} y_{p} & =A\cos \left (nx\right ) +B\sin \left (nx\right ) \\ & =\frac {\sin \left (nx\right ) }{1-n^{2}} \end {align*}
Which is the same as variation of parameters method.
Note: Full solution should also really consider the case for \(n=1\). Will update later.