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Thread: 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:

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

    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:
    $\displaystyle a x^2 y'' + bxy' +cy=0$, 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:
    $\displaystyle a x^2 y'' + bxy' +cy=0$, 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
    $\displaystyle u'' - \frac{1}{x}u' - 4x^2u=0$
    put $\displaystyle x=\sqrt{t}.$ then $\displaystyle u'=2\sqrt{t} \frac{du}{dt},$ and $\displaystyle u''=2\frac{du}{dt} + 4t \frac{d^2u}{dt^2}.$ then your differential equation becomes: $\displaystyle \frac{d^2u}{dt^2}-u=0,$ which is very easy to solve.

    $\displaystyle u''' + 2e^xu'' -u' -2e^xu=0$
    put $\displaystyle u'' - u=v.$ then your equation becomes: $\displaystyle v'+2e^xv=0,$ which clearly has $\displaystyle v=e^{-2e^x}$ as a solution. so now you need to solve $\displaystyle u''-u=e^{-2e^{x}}.$
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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
    $\displaystyle x^2 y''(x) + xy'(x)-n^2y(x)=0$
    Euler type equations have polynomial type solutions.
    Just substitute $\displaystyle y(x)=x^{\lambda}$ and get a parabola of $\displaystyle \lambda$, solve it and get two roots $\displaystyle \lambda_{1}$ and $\displaystyle \lambda_{2}$ (this is for second-order equations, for higher-order equations you get polynomial of degree $\displaystyle n$).

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

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