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The Problem

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What I Understand I understand how to find a general solution to this equation. The homogeneous general solution is:

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However, it has been a while since I have solved for the particular solution. How would I set this part up?

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up vote 1 down vote accepted

Attempt a solution of the form

$$y_p(t) = (At+B)te^{3t}.$$

Differentiating we get

$$y_p' (t) = 2Ate^{3t} + 3At^2 e^{3t} + Be^{3t} + 3Bt e^{3t},$$

$$y_p ' ' (t) = 2Ae^{3t} +12At e^{3t} +9At^2 e^{3t} +6B e^{3t} +9 Bt e^{3t}.$$

Substituting into $y_p ' ' - 9 y_p = t e^{3t}$ we obtain

$$2A e^{3t} +12 A t e^{3t} +6B e^{3t} = (2A + 6B) e^{3t} + 12A t e^{3t} \equiv t e^{3t}.$$

Equating coefficients leads to the system

$$\begin{cases} 2A + 6B & = 0, \\ 12A & = 1, \end{cases}$$

with solution

$$A = \frac{1}{12}, B = - \frac{A}{3} = - \frac{1}{36}.$$


$$y(t) = c_1 e^{3t} + c_2 e^{-3t} + \frac{t^2 e^{3t}}{12} - \frac{t e^{3t}}{36}.$$

I hope this clears it up. I had to actually sit down and go through all of it (many times) to be certain it was correct.

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I would have to multiply through by "t" to make sure that the particular solution didn't match up with the general solution right? – user2738319 Dec 10 '13 at 2:36
@user2738319 Yes, correct. I forgot that. Thank you! – Mark Fantini Dec 10 '13 at 2:38
Whenever I attempt to solve the system of equation I get A and B equal to zero. This is not correct or is it? – user2738319 Dec 10 '13 at 3:07
@user2738319 I've edited it and put in all steps. Check it now. – Mark Fantini Dec 10 '13 at 16:59

Hint: Choose

$$y_p(t) = t e^{3t}(a + b t)$$

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