# Undetermined Coefficients

I tried to change it to $y" + y = (14\sin x-28 \sin^3 x)$. The complementary solution is $C_1\cos x+C_2\sin x$ and the particular solution to $y" + y = 14 \sin x$ is $-7\sin x$. How do you find the particular solution to $y" + y = -28\sin^3 x$ ? What would be your guess?

Ans: $$c_1 \sin(x\sqrt2)+c_2\cos(s\sqrt2)-7\sin x-\sin(3x)$$

• You have the right idea, you need to convert $\sin(x)\cos(2x)$ into terms of sines and cosines. – Chee Han Feb 12 '17 at 6:58

Using double-angle formula, you can rewrite $\sin(x)\cos(2x)$ as $$\sin(x)\cos(2x) = \frac{1}{2}[\sin(3x) - \sin(x)].$$ Indeed, \begin{align*} \sin(3x) = \sin(2x + x) & = \sin(2x)\cos(x) + \cos(2x)\sin(x) \\ \sin(x) = \sin(2x-x) & = \sin(2x)\cos(x) - \cos(2x)\sin(x). \end{align*} Thus, this gives $$y'' + y = 7[\sin(3x) - \sin(x)].$$ Since there is no first derivative involved, normally one would guess a particular solution of the form $$y_p(x) = A\sin(3x) + B\sin(x).$$ Since $\sin(x)$ is a solution to the homogeneous equation $y''+y=0$, one should now guess a particular solution of the form $$y_p(x) = A\sin(3x) + Bx\sin(x) + Cx\cos(x).$$