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

**vitaemage** · August 11th 2008, 10:14 AM

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

General Solution
$r^2 + 3r + 2 = 0$
$(r + 1)(r + 2) = 0$
$r = -1, -2$
$y = c1*e^{-1x} + c2*e^{-2x} + yparticular$

Particular Solution
$yparticular = A*e^{-2x}$
$yparticular' = -2A*e^{-2x}$
$yparticular'' = 4A*e^{-2x}$
$4A*e^{-2x} -6A*e^{-2x}+2A*e^{-2x} = 5*e^{-2x}$
$6A-6A = 5$
$0 = 5$

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

**mr fantastic** · August 11th 2008, 12:59 PM

Originally Posted by vitaemage

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

General Solution
$r^2 + 3r + 2 = 0$
$(r + 1)(r + 2) = 0$
$r = -1, -2$
$y = c1*e^{-1x} + c2*e^{-2x} + yparticular$

Particular Solution
$yparticular = A*e^{-2x}$
$yparticular' = -2A*e^{-2x}$
$yparticular'' = 4A*e^{-2x}$
$4A*e^{-2x} -6A*e^{-2x}+2A*e^{-2x} = 5*e^{-2x}$
$6A-6A = 5$
$0 = 5$

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 $y_p = a x e^{-2x}$ .

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

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