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Math Help - general Solution to a Linear, Homogenious Differential Equation

  1. #1
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    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
    but I have no idea how to approach the problem from here. Can somebody please help?
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  2. #2
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    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?
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  3. #3
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    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?
    Last edited by nivekious; September 16th 2010 at 02:05 PM.
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  4. #4
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    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?
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  5. #5
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    Oh I think I see, would it be u*x? so u'=xe^(x^2)*u?
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  6. #6
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    Don't forget the constant of integration.
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  7. #7
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    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?
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  8. #8
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    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?
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  9. #9
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    Yes, thank you so much for your help.
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  10. #10
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    You're welcome. Have you got it from here on out?
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  11. #11
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    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?
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  12. #12
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    Well, you can, I suppose. I'd probably try the integrating factor method first. That works on first-order linear non-homogeneous equations, right?
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  13. #13
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    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.
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  14. #14
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    Great! You're very welcome. Have a good one!
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