# Thread: Second order ODE problem

1. ## 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.

2. 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.

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