# Parameter variation method.

#### Ale_M

Can someone help me with this problem? y"-y=xe^× y(0)=1 y'(0)=0

#### topsquark

Math Team
Can someone help me with this problem? y"-y=xe^× y(0)=1 y'(0)=0
First solve $$\displaystyle y_h ^{ \prime \prime } - y_h = 0$$ (The homogeneous solution.)

Then use variation of parameters for the particular solution. I'd start with $$\displaystyle y_p = Ax^2 e^x + B x e^x$$
(You could use $$\displaystyle y_p = A x^2 e^x + B x e^x + C e^x$$ but the C term cancels out anyway.)

-Dan

idontknow