
Second order ODE problem
$\displaystyle y''  3y' + 2y = 5e^{2x}$
Homogenous solution is:
$\displaystyle Ae^{2x} + Be^x$
Particular solution guess:
$\displaystyle y = axe^{2x}$
$\displaystyle y' = \alpha e^{2x}(2x + 1)$
$\displaystyle y'' = 4\alpha e^{2x}(x + 1)$
Subbing into ODE:
$\displaystyle 4\alpha e^{2x}(x + 1)  3\alpha e^{2x}(2x + 1) + 2\alpha xe^{2x} = 5e^{2x}$
My text says this simplifies to:
$\displaystyle 4\alpha e^{2x}  3\alpha e^{2x} = 5e^{2x}$
How is this so? I understand that you can make this true by making x = 0, but wouldn't this make the RHS = 5?
I'm sure I'm confusing myself, any help would be great.

You need to expand and collect like terms...
$\displaystyle \displaystyle 4\alpha e^{2x}(x + 1)  3\alpha e^{2x}(2x + 1) + 2\alpha x\,e^{2x} = 5e^{2x}$
$\displaystyle \displaystyle 4\alpha x\,e^{2x} + 4\alpha e^{2x}  6\alpha x\,e^{2x}  3\alpha e^{2x} + 2\alpha x\,e^{2x} = 5e^{2x}$.
The $\displaystyle \displaystyle x\,e^{2x}$ terms all cancel.

Ahh right. I thought they factored it because there was some convenient way to solve it. I guess not! Thanks.