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

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  • $\begingroup$ You have the right idea, you need to convert $\sin(x)\cos(2x)$ into terms of sines and cosines. $\endgroup$ – Chee Han Feb 12 '17 at 6:58
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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 $$ y_p(x) = A\sin(3x) + Bx\sin(x) + Cx\cos(x). $$

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  • $\begingroup$ cos 2x = cos2 x – sin2 x or cos 2x = 2 cos2 x – 1 or cos 2x = 1 – 2 sin2 x are the only trig identities $\endgroup$ – Shay Feb 12 '17 at 7:16
  • $\begingroup$ You mean you are only allowed to use those identities? Otherwise, en.wikipedia.org/wiki/List_of_trigonometric_identities $\endgroup$ – Chee Han Feb 12 '17 at 7:28
  • $\begingroup$ where did the 14 go? $\endgroup$ – Shay Feb 12 '17 at 7:34
  • $\begingroup$ Opps forgot about the 14, just added that. $\endgroup$ – Chee Han Feb 12 '17 at 7:35
  • $\begingroup$ The answers shown but if I do it your way I get -9Asin3x-Bsinx+Asin3x+Bsinx=14sin3x-14sinx and I get -7/2 sin3x $\endgroup$ – Shay Feb 12 '17 at 7:45

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