# Thread: Diff Eq. Non-homogeneous const coeffs, particular solution is particularly hard

1. ## Diff Eq. Non-homogeneous const coeffs, particular solution is particularly hard

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

2. Originally Posted by 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
You should know to try $\displaystyle y_p = a x e^{-2x}$.

This 'problem' occurs whenever $\displaystyle \lambda = \lambda_1$ is one of the solutions to the auxillary equation and $\displaystyle e^{\lambda_1 x}$ is one of the terms in the right hand side.