Undetermined coefficient guessing the particular solution for $y^{"}+k^2y=\sin(kx)$.

Question

since the homogeneous solution also has the same form as the nonhomogeneous part I was a bit confused on how to guess the form of the particular solution. any help is greatly appreciated. My guess is the following $$Y_P =C \sin(kx) +D \cos(kx)$$ but I think the correct guess would be $Y_P=(C+Dx)(A\sin(kx) + B\cos(kx))$ where the $A\sin(kx)+B\cos(kx)$ is our homogeneous solution.

Answer 1

As you mentioned the homogeneous solution is

$${y_h} = A\cos (kx) + B\sin (kx)\tag{1}$$

and hence your particular solution $y_p$ surely cannot be like $(1)$ as it just make the LHS of the ODE being zero, leaving you with nothing! :)

So we may try particular solutions of the form

$${y_p} = Cx\sin (kx) + Dx\cos (kx)\tag{2}$$

Put it into the ODE to obtain

$$\begin{array}{l} C\left[ {(2k\cos (kx) - {k^2}x\sin (kx) + {k^2}x\sin (kx)} \right] + \\ D\left[ {( - 2k\sin (kx) - {k^2}x\cos (kx) + {k^2}x\cos (kx)} \right]\\ = 2Ck\cos (kx) - 2Dk\sin (kx) = \sin (kx) \end{array}\tag{3}$$

and hence

$$\left\{ \begin{array}{l} C = 0\\ D = - \frac{1}{2k} \end{array} \right.\tag{4}$$

and finally your particular solution is

$${y_P} = - \frac{1}{2k}x\cos (kx)\tag{5}$$

and your general solution will be

$$y = {y_h} + {y_p} = A\cos (kx) + B\sin (kx) - \frac{1}{2k}x\cos (kx)\tag{6}$$

Answer 2

Your first guess is the general solution of the homogeneous equation. The correct guess is what you believe to be correct, where you can take $C=0$ and $D=1$.