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Math Help - 2nd Order ODE

  1. #1
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    2nd Order ODE

    Any ideas how can I solve this equation f''=a*f^0.4?
    a=real number

    Thank you
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  2. #2
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    Re: 2nd Order ODE

    I'm not certain on this but I did the following:




    Dealing with each side of the equation individually:

    [IMG]http://latex.codecogs.com/gif.latex?\displaystyle \begin{align*} \int f'(x)f''(x)dx \\ &= \int f''(x)dx \times f'(x)-\iint f''(x)dx \times Dx(f'(x)) \\ &= f'(x)^2 - \int f'(x)f''(x)dx \\ \\ \Rightarrow \int f'(x)f''(x)dx=\frac{1}{2}f'(x)^2 \end{align*}[/IMG]

    And

    [IMG]http://latex.codecogs.com/gif.latex?\begin{align*} a \int f'(x)f(x)^{0.4}dx \\ &=a\left ( \int f'(x)dx\times f(x)^{0.4}-\iint f'(x)dx \times Dx(f(x)^{0.4})dx \right ) \\ &=a\left ( f(x)^{1.4}-\int f(x) \times 0.4f'(x)f(x)^{-0.6}dx \right ) \\ &=a\left ( f(x)^{1.4}-0.4\int f'(x)f(x)^{0.4}dx \right ) \\ \Rightarrow a \int f'(x)f(x)^{0.4}dx = \frac{5a}{7}f(x)^{1.4} \end{align*}[/IMG]

    Giving:

    [IMG]http://latex.codecogs.com/gif.latex?\displaystyle \begin{align*} \frac{1}{2}f'(x)^2 = \frac{5a}{7}f(x)^{1.4} \end{align*}[/IMG]

    Integrating one more time should be relatively straightforward now. But please don't work straight off this answer - Do everything above and make sure I haven't made a mistake - it's 2am right now. =D

    EDIT: Sorry, I can't get the latex to display properly - just copy/paste everything between the image tags into your browser and you'll see the equations. If anyone can tell me why the equations aren't displaying I'd be grateful.
    Last edited by Ivanator27; July 16th 2012 at 04:32 PM.
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  3. #3
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    Re: 2nd Order ODE

    Quote Originally Posted by theod9 View Post
    Any ideas how can I solve this equation f''=a*f^0.4?
    a=real number

    Thank you
    \displaystyle \begin{align*} \frac{d^2f}{dx^2} &= a\,f^{0.4} \\ \frac{df}{dx}\,\frac{d^2f}{dx^2} &= a\,f^{0.4}\,\frac{df}{dx} \end{align*}

    Now if we let \displaystyle \begin{align*} u = \frac{df}{dx} \end{align*}, then we can write

    \displaystyle \begin{align*} u\,\frac{du}{dx} &= a\,f^{0.4}\,\frac{df}{dx} \\ \int{u\,\frac{du}{dx}\,dx} &= \int{a\,f^{0.4}\,\frac{df}{dx}\,dx} \\ \int{u\,du} &= \int{a\,f^{0.4}\,df} \\ \frac{u^2}{2} + C_1 &= \frac{a\,f^{1.4}}{1.4} + C_2 \\ \frac{u^2}{2} &= \frac{5a\,f^{1.4}}{7} + C_2 - C_1 \\ u^2 &= \frac{10a\,f^{1.4} + C}{7} \textrm{ where } C = 14\left(C_2 - C_1\right) \\ u &= \pm \sqrt{\frac{10a\,f^{1.4} + C}{7}} \\ \frac{df}{dx} &= \pm \frac{\sqrt{10a\,f^{1.4} + C}}{\sqrt{7}} \\ \pm \frac{\sqrt{7}}{\sqrt{10a\,f^{1.4} + C}}\,\frac{df}{dx} &= 1 \\ \int{\pm \frac{\sqrt{7}}{\sqrt{10a\,f^{1.4} + C}} \, \frac{df}{dx}\,dx} &= \int{1\,dx} \\ \int{\pm \frac{\sqrt{7}}{\sqrt{10a\,f^{1.4} + C}}\,df} &= x + C_3  \end{align*}

    To solve this integral you need to use the Hypergeometric Function as shown here.
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    Re: 2nd Order ODE

    Quote Originally Posted by Ivanator27 View Post
    \displaystyle \begin{align*} \frac{1}{2}f'(x)^2 = \frac{5a}{7}f(x)^{1.4} \end{align*}

    Integrating one more time should be relatively straightforward now. But please don't work straight off this answer - Do everything above and make sure I haven't made a mistake - it's 2am right now. =D
    Sorry but this equation here is missing an integration constant, which makes the integral NOT straightforward.
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  5. #5
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    Re: 2nd Order ODE

    Ah - stupid of me to forget that. My apologies.

    On another note - any idea why my latex didn't display properly?
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