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For an odd integrable function, both $f(x)$ and $f(x)\cos nx$ are odd. Their integrals over $[-\pi ,\pi ]$ therefore vanish, so the constant and cosine Fourier coefficients are zero. For $f(x)=\sin ax$, the remaining Fourier sine coefficients areThe product-to-sum identity in the first step reduces the integral to two cosines. Here $a$ is real and noninteger, so no denominator vanishes. Consequently the Fourier series of a noninteger-frequency sine isAt $x=\pm \pi$ every term is zero. The Fourier series converges there to the average of the two limits of the periodic extension, namely zero, rather than to the nonzero endpoint values $\pm \sin \pi a$.

Use the method of undetermined coefficients. Substitution of $y_p=(a+bx)\sin x+(c+dx)\cos x$ givesEquating the coefficients of $x\sin x$, $x\cos x$, $\sin x$, and $\cos x$ to the forcing gives $d=1$, $b=0$, $c=1$, $a=-1$. ThusThe characteristic equation of the homogeneous equation is $(r-1{)}^2=0$, with a repeated root. Its complementary function is $(C_1+C_{2}x)e^x$. Since $y_p(0)=1$ and $y_p^{\prime }(0)=0$, the initial conditions give $C_1=-1$ and $C_2=1$. ThereforeDirect differentiation verifies both zero initial data and the original forcing.