Instead of differentiating on the LHS, try to see what's going on in the RHS. Can you write the RHS as a total derivative? If so, what can you do then?
I need to solve d/dx(e^(-(x^2))*(du/dx))=u+x*(du/dx)
I get e^(-(x^2))*u''-2xe^(-(x^2))*u'=u+x*u'
u''-2x*u'=e^(x^2)*u+xe^(x^2)*u'
u''+(2x-xe^(x^2))*u'-e^(x^2)*u=0
but I have no idea how to approach the problem from here. Can somebody please help?
Thanks, but I'm not sure what you mean by total derivative. I looked it up but found several definitions and I'm not sure which would apply to this scenario. Could you please clarify what you mean? I assume that what you're getting at would eventually lead to integrating both sides to get rid of the d/dx on the left, but if I were to do that in the original state I would just wind up with e^(-(x^2))*u' = ∫u+x*u. Should I do this and then substitute in v=∫u and solve for v, or am I way off?
I believe your computations are correct so far. The fact that your equation is no longer homogeneous shouldn't be too much of a concern. You started out with a homogeneous second-order ODE (as shown by your computations in the OP). When you integrate once, you're going to have an arbitrary constant, making the next integration to be that of a DE that is not homogeneous. It's the way of any second-order DE that you solve by doing two integrations sequentially (instead of just writing the solution down, like with some second-order DE's). The only way you would not have a homogeneous DE right now is if you could show by using an initial condition that the constant is zero. Make sense?