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**vitaemage** $\displaystyle y'' + 3y' + 2y = 5e^{-2x}$ <-- problem

General Solution

$\displaystyle r^2 + 3r + 2 = 0 $

$\displaystyle (r + 1)(r + 2) = 0$

$\displaystyle r = -1, -2$

$\displaystyle y = c1*e^{-1x} + c2*e^{-2x} + yparticular$

Particular Solution

$\displaystyle yparticular = A*e^{-2x}$

$\displaystyle yparticular' = -2A*e^{-2x}$

$\displaystyle yparticular'' = 4A*e^{-2x}$

$\displaystyle 4A*e^{-2x} -6A*e^{-2x}+2A*e^{-2x} = 5*e^{-2x}$

$\displaystyle 6A-6A = 5$

$\displaystyle 0 = 5$

The conclusion I arrive at after figuring out the particular solution cannot be true and I am lost as to what this means and or how to solve the problem. Can anyone help clear this up for me?

Thanks in advance