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Given $y''' - 5y'' - y' + 5y = 3e^{-x}$, find the general solution.

I found the roots for the homogeneous solution to be 5, 1, and -1:

$$(r - 5)(r + 1)(r - 1)=0$$

$$y_h(x) = c_1e^x + c_2e^{-x} + c_3e^{5x}$$

Setting up the particular solution, I have:

$$g(x) = 3e^{-x}$$

$$y_p(x) = Axe^{-x}$$

$$y_p'(x) = Ae^{-x} - Axe^{-x} = A(1 - x)e^{-x}$$

I know I need to use the product rule to continue differentiating $y'$, but is there an easier method to do so?

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    If you're doing differential equations like this, shouldn't you have learned how to take a derivative long ago?2012-11-28

3 Answers 3