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Math Help - I need help with 3 ODEs.

  1. #1
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    I need help with 3 ODEs.

    I'm stuck with these three ODEs:

    u'' - \frac{1}{x}u' - 4x^2u=0
    x^2 y''(x) + xy'(x)-n^2y(x)=0
    u''' + 2e^xu'' -u' -2e^xu=0

    Nothing I tried worked, could you please help me? A hint on which method to try would be really great.

    Thank you!
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  2. #2
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    The second one is Bessel's equation, I think. I'll Google for the solution.
    If anyone has an idea for the other two..
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by marianne View Post
    The second one is Bessel's equation, I think. I'll Google for the solution.
    If anyone has an idea for the other two..
    i would say the second was an Euler equation. have you considered series solutions for the other two?
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  4. #4
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    Quote Originally Posted by Jhevon View Post
    i would say the second was an Euler equation. have you considered series solutions for the other two?
    I've found that Euler equation has the form:
    a x^2 y'' + bxy' +cy=0, but I don't have a constant times y, but something dependent on x..?

    Edit: I thought you were talking about the first one, sorry. :-(

    Also, we haven't done series solutions in class yet, but I'll try to Google it. Thank you.
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by marianne View Post
    I've found that Euler equation has the form:
    a x^2 y'' + bxy' +cy=0, but I don't have a constant times y, but something dependent on x..?

    Also, we haven't done series solutions in class yet, but I'll try to Google it. Thank you.
    -n^2 is a constant as far as a function of x is concerned

    i will think of other ways to do the problems. if you haven't done series solutions, it would be a lot to learn just to do these. so you probably aren't expected to do them that way
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  6. #6
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    Quote Originally Posted by Jhevon View Post
    -n^2 is a constant as far as a function of x is concerned
    Yes, I know, I edited the post above, I thought you were talking about the first one. It's not an excuse, but it's late in this timezone. :-)

    i will think of other ways to do the problems. if you haven't done series solutions, it would be a lot to learn just to do these. so you probably aren't expected to do them that way
    Oh, thank you! And I will read about series solutions, maybe they aren't that hard to learn.
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  7. #7
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    I solved the second one. In case someone has a similar problem, I've found this page Pauls Online Notes : Differential Equations - Euler Equations to be useful.
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  8. #8
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    Quote Originally Posted by marianne View Post
    u'' - \frac{1}{x}u' - 4x^2u=0
    put x=\sqrt{t}. then u'=2\sqrt{t} \frac{du}{dt}, and u''=2\frac{du}{dt} + 4t \frac{d^2u}{dt^2}. then your differential equation becomes: \frac{d^2u}{dt^2}-u=0, which is very easy to solve.

    u''' + 2e^xu'' -u' -2e^xu=0
    put u'' - u=v. then your equation becomes: v'+2e^xv=0, which clearly has v=e^{-2e^x} as a solution. so now you need to solve u''-u=e^{-2e^{x}}.
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  9. #9
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    Thank you so much!!!
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  10. #10
    Senior Member bkarpuz's Avatar
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    Exclamation Euler type.

    Quote Originally Posted by marianne View Post
    x^2 y''(x) + xy'(x)-n^2y(x)=0
    Euler type equations have polynomial type solutions.
    Just substitute y(x)=x^{\lambda} and get a parabola of \lambda, solve it and get two roots \lambda_{1} and \lambda_{2} (this is for second-order equations, for higher-order equations you get polynomial of degree n).

    Then, the solution is
    <br />
y(x)=<br />
\begin{cases}<br />
c_{1}x^{\lambda_{1}}+c_{2}x^{\lambda_{2}},&\lambda  _{1}\neq\lambda_{2}\\<br />
x^{\lambda_{1}}\big[c_{1}+c_{2}\ln(x)\big],&\lambda_{1}=\lambda_{2},<br />
\end{cases}<br />
    where c_{1} and c_{2} are arbitrary constants.

    Note. Note that \bigg(\frac{d}{dx}\bigg)^{k} will decrease the power of y(x)=x^{\lambda} by k while the factor x^{k} will increase it by k, hence you will get x^{k}\bigg(\frac{d}{dx}\bigg)^{k}x^{\lambda}=\lamb  da(\lambda-1)\cdots(\lambda-k+1)x^{\lambda} for any k\in\mathbb{N}.
    This is why we search solutions of polynomial type.
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