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Solve the following ODE using the method of undetermined coefficients:

$$\frac{d^2y}{dx^2}+9y=x+\cos x\sin(2x)$$

I know how to go about solving problems like this, the complementary function or the $y(\text{general})$ for this question would be

$$y(g) = C_1 \cos(\sqrt 3 x)+ C_2 \sin(\sqrt 3 x)$$

I'm having trouble determining the particular equation to go about solving it. As for the $x$ it can be written as, $Ax+B$ whereas the second part, I'm not sure what $y(p)$ would be.

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Your complementary solution has an error. The $\sqrt{3}$'s should just be $3$'s.

You're right about $Ax+B$. For the other piece, I would use the product to sum trig identities, one of which is $$\sin x \cos y = \frac{\sin(x+y)+\sin(x-y)}{2}.$$

In this case you'd get $$\frac{\sin{3x}+\sin(x)}{2}.$$

The particular solution will be a bit scary. The $\sin x$ is easy, but note that the $\sin 3x$ matches a complementary solution.

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  • $\begingroup$ Could you explain why the sin and cos x in the complementary solution be 3? Wouldn't alpha be 0 and beta be √3 if u were to right √-9 as √3 *√-3 ? $\endgroup$ – Gary Andrews30 Apr 10 '17 at 12:10
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    $\begingroup$ The equation is $r^2+9 = 0$ so $r=\pm \sqrt{9} =\pm 3$. Why are you taking the square root twice? $\endgroup$ – B. Goddard Apr 10 '17 at 12:14
  • $\begingroup$ isnt one component of the force is the solution to hgs problem. so you will have to look for a solution of the form $Ax\cos 3x.$ $\endgroup$ – abel Apr 10 '17 at 19:22
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I know how to go about solving problems like this, the complementary function or the y(general) for this question would be y(g) = C1 cos(√3 x)+ C2 sin(√3 x)

Where do the square roots come from? You should find: $$y_c(x) = c_1\cos(3x)+c_2\sin(3x)$$

As for the x it can be written as, Ax+B

Correct!

whereas the second part, I'm not sure what y(p) would be.

Note that using trig formulas, you have: $$\cos (x) \sin (2x) = \frac{1}{2}\left( \sin(x) + \sin(3x) \right)$$ Does that help? Careful with the $3x$-argument. Can you take it from there?

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  • $\begingroup$ Could you explain why the sin and cos x in the complementary solution be 3? Wouldn't alpha be 0 and beta be √3 if u were to right √-9 as √3 *√-3 ? $\endgroup$ – Gary Andrews30 Apr 10 '17 at 12:12
  • $\begingroup$ @GaryAndrews30 The characteristic equation for $y''+9y=0$ is $r^2+9=0$ and this should lead to $\cos(3x)$ and $\sin(3x)$. I'm guessing your formula for $\beta$ should give you $\sqrt{9} = 3$... $\endgroup$ – StackTD Apr 10 '17 at 12:15

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