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\begin{cases} u''+a^2u=\sin(\pi x),\quad 0

I found the characteristic equation to be $$r^2+a^2=0\implies r=\pm ai$$

So $$u=C\cos(ax)+D\sin(ax)$$ $$u'=-Ca\sin(ax)+Da\cos(ax)$$ $$u''=-Ca^2\cos(ax)-Da^2\sin(ax)$$

I find substitute these expressions into the BVP to get,

$$(-Ca^2\cos(ax)-Da^2\sin(ax))+a^2(C\cos(ax)+D\sin(ax))=\sin(\pi x),$$

But I'm just getting the left side terms cancelling themselves out...

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    Right! With the characteristic equation you solved the homogeneous equation (when the right side is zero). To get a complete solution, you have to add a special solution of the inhomogeneous equation which can be obtained by an ansatz using harmonic functions. – 2017-02-22
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    Of course! $u=u_h+u_p$. You are saying I need to find $u_p$, but will this be in the form $u_p=u=C\cos(\pi x)+D\sin(\pi x)$? Find the second derivative of $u_p$ and then $u''=u_h''+u_p''$? So in the end I should have $u_h''+u_p'' + a^2(u_h+u_p)=sin(pi x)$ and then I would plug in our initial conditions? – 2017-02-22
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    Perhaps the inhomogenous solution (or particular solution, special solution, etc.) is of the form $u_I(x) = Ax\sin{(ax)} + Bx\sin{(ax)}.$ – 2017-02-22
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    Would I then simply change the BVP to $(u_H+u_I)''+a^2(u_H+u_I)=\sin(\pi x)$? – 2017-02-22

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The general solution of the homogeneous equation is as you found out $$ u_h(x)=A\cos(a\,x)+B\sin(a\,x). $$ To find a particular solution you can use the method of variation of constants or the method of undetermined coefficients, which in this case is simpler. If $a\ne\pi$, a particular solution is of the form $$ u_p(x)=C_1\cos(\pi\,x)+C_2\sin(\pi\,x). $$ Plug this into the equation to find $C_1$ and $C_2$.

If $a=\pi$, then the particular solution will be of the form $$ u_p(x)=x\bigl(C_1\cos(\pi\,x)+C_2\sin(\pi\,x)\bigr). $$

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    Would there be a more general form for all $a\in\mathbb{R}$? Perhaps the latter form that you have provided? If it is, I am wondering why the x is multiplied to both $\cos$ and $\sin$. – 2017-02-22
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    When $a=\pi$ the $x$ is needed because in that case $\sin(\pi\,x)$ is a solution of the homogenous equation. In general, both $\cos$ and $\sin$ must appear, even if the "independent" term has only one of them. – 2017-02-23