(adsbygoogle = window.adsbygoogle || []).push({}); Diff. Eq. Problem ---> y'' - 2y' - 3y = -3 t e^(-t) What is general solution?

This is problem number 3 in section 3.6 in Boyce, DiPrima 8th Edition "Elementary Differential Equations and Boundry Value Problems"

Find the general solution of the differential equation:

[tex]y'' - 2y' - 3y = -3 t e^{-t}[/tex]

Here is what I have done so far:

The characteristic equation is [tex]r^2 - 2r - 3 = 0[/tex] and [tex]r_1 = 3, r_2 = -1[/tex]

That gives [tex]y(t) = C_1 e^{3t} + C_2 e^{-t}[/tex] as the complementary solution.

Now I run into trouble when selecting the superposition equation for using the method of undetermined coefficients. I tried [tex]Y(t) = (A t + B) e^{-t}[/tex] with no luck. I get [tex]A = \frac{3}{4} t[/tex] for an answer, which is incorrect.

The book lists the answer as [tex]y(t) = C_1 e^{3t} + C_2 e^{-t} + \frac{3}{16} t e^{-t} + \frac{3}{8} t^2 e^{-t}[/tex]

What am I doing wrong? Is the [tex]Y(t)[/tex] I chose correct? Please help!

Thank you.

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# Homework Help: Diff. Eq. Problem -> y'' - 2y' - 3y = -3 t e^(-t) What is general solution?

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