# Differential Equation

• Feb 6th 2011, 04:55 AM
BabyMilo
Differential Equation
$\frac{d^2y}{dx^2}+\frac{dy}{dx}-2y=-3e^x$

given y=2 when x=0, and y=0 when x=2

I found that $y= Ae^x+Be^{-2x}-3xe^x$

then $2=A+B$
and $0=Ae^2+Be^{-4}-6e^2$

is this correct so far?

thanks
• Feb 6th 2011, 07:13 AM
FernandoRevilla
Quote:

Originally Posted by BabyMilo
I found that $y= Ae^x+Be^{-2x}-3xe^x$

Right.

Quote:

then $2=A+B$
and $0=Ae^2+Be^{-4}-6e^2$

Right with those initial conditions but I guess they could be something like this:

$y(0)=2,\quad y'(0)=2$

Check the conditions.

Fernando Revilla