show y= Ax^2e^x is a solution to:

[calchere]

Show that y=Ax^2e^x is the general solution to the differential equation:
y'' - y' = 2e^x(x+1)

y= Ax^2e^x
y'= 2Axe^x = 2xe^x
y''= 2Ae^x = 2e^x

(2e^x)-(2xe^x) = 2e^x(x+1)
2e^x - 2xe^x = 2xe^x + 2e^x

Is this correct? Then it wouldn't be a solution to the differential equation?

 

[stapel]

y = Ax^2e^x
y'= 2Axe^x = 2xe^x

How are you getting that 2Axe^x is the derivative of Ax²e^x? At the very least, the Product Rule says that the derivative should have two terms.

Also, how are you getting that 2Axe^x equals 2xe^x? Were you given that A = 1? If so, why not simplify at the beginning...?

Thank you.

Eliz.

 

[calchere]

Ok, i didn't even realize i had to use the product rule. Most of my other problems are only one term.
Thank you.