Solving Linear ODE: dy/dx-2y=x^2e^2x

[NewtonianAlch]

Homework Statement

Solve [itex]\frac{dy}{dx}[/itex] - 2y = x[itex]^{2}[/itex]e[itex]^{2x}[/itex]

The Attempt at a Solution

Integrating factor = e[itex]^{2x}[/itex]

So we multiply through the given equation by the integrating factor and get:

e[itex]^{2x}[/itex][itex]\frac{dy}{dx}[/itex] - 2e[itex]^{2x}[/itex]y = x[itex]^{2}[/itex]e[itex]^{4x}[/itex]

Contract the left-hand side via the chain rule to get:

[itex]\frac{d}{dx}[/itex](e[itex]^{2x}[/itex]y) = x[itex]^{2}[/itex]e[itex]^{4x}[/itex]

Integrate both sides

e[itex]^{2x}[/itex]y = [itex]\frac{1}{32}e^{4x}[/itex](8x[itex]^{2}[/itex]-4x+1)+C

Now divide through by e[itex]^{2x}[/itex] and the equation definitely does not equal what Wolfram Alpha gives as the solution:

y = [itex]\frac{1}{3}[/itex]e[itex]^{2x}[/itex]x[itex]^{3}[/itex]+Ce[itex]^{2x}[/itex]

I checked some of the parts individually with Wolfram, such as the integration of the right-hand side and that was correct, so I'm not too sure what's causing the difference in answers.

 

[NewtonianAlch]

Grr, nvm. I forgot the minus sign in the integrating factor -.-