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The question I am stuck on is:

Find a particular solution to

$y'' + 36y = 36\sin(6t)$

I guessed that $y_p(t)$ would be $A\sin(6t)$, because I would be taking the derivative twice, so cosine would not come into play. So I used the method of undetermined coefficients to try and find $A$. However, I hit a bit of a roadblock.

When I went through my math, I got:

$$-36A\sin(6t) + 36A\sin(6t) = 36\sin(6t).$$

A can't be zero as far as I can tell.

I'm wondering what the answer would be for such a case (if this is even a special case) or if I did something wrong in my math.

Thank you in advance for any help you can give.

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    $\begingroup$ Guess $y_p = t(a \cos 6 t + b \sin 6 t)$ and solve for the coefficients. Because the homogeneous solution includes $\cos(6 t) + b \sin(6 t)$, we multiply by that extra $t$. See Method of Undetermined Coefficients. See the link I added - especially the examples and table. It is a common problem! $\endgroup$ – Amzoti Mar 10 '15 at 20:22
  • $\begingroup$ How did you infer that? Is this indeed a special case or is this a common problem? Ah! Thank you. I guess it would be safe to say I should memorize all of those for exams. $\endgroup$ – Kommander Kitten Mar 10 '15 at 20:23
  • $\begingroup$ You bet! Moreover - practice many examples to get accurate results! $\endgroup$ – Amzoti Mar 10 '15 at 20:26
  • $\begingroup$ By the way, you will get $a = -3, b = 0$ when solving. $\endgroup$ – Amzoti Mar 10 '15 at 20:31

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