# Thread: general Solution to a Linear, Homogenious Differential Equation

1. ## general Solution to a Linear, Homogenious Differential Equation

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

2. 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?

3. 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?

4. I don't think you're quite getting the idea. The LHS is written as a derivative of a function. It turns out you can write the RHS in the same fashion:

$\displaystyle{u+xu'=\frac{d}{dx}(?)}$

What takes the place of the question mark?

5. Oh I think I see, would it be u*x? so u'=xe^(x^2)*u?

6. Don't forget the constant of integration.

7. Hmm wouldn't that leave me with u'-xe^(x^2)*u=ce^(x^2), making it nonhomogenious? Or am I not putting the constant in the right place?

8. 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?

9. Yes, thank you so much for your help.

10. You're welcome. Have you got it from here on out?

11. I think so, I need to get the general solution of u'-xe^(x^2)*u=0 and add it to a particular solution of u'-xe^(x^2)*u=ce^(x^2), right?

12. Well, you can, I suppose. I'd probably try the integrating factor method first. That works on first-order linear non-homogeneous equations, right?

13. Ah, forgot about that, probably would have saved some time. I did it now and I got the same solution both ways though, so it all worked out. Thanks again.

14. Great! You're very welcome. Have a good one!