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Math Help - Solving an IVP for y(x)

  1. #1
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    Solving an IVP for y(x)

    \frac{d^2y}{dx^2} = \frac{1}{a} \sqrt{1 + {(\frac{dy}{dx})^2}} ; y(0)=a>0, y'(0)=0

    let w = y'

    \frac{dw}{\sqrt{1+w^2}} = \frac{1}{a}dx

    ln|w + \sqrt{w^2 + a} | = \frac{x}{a} + c

    ln|y' + \sqrt{y'^2 + a} | = \frac{x}{a} + c

    for the given values c = 0

    y' + \sqrt{y'^2 + a} = e^\frac{x}{a}

    From here I don't know how to proceed. How do you reduce it such that I can get y(x) = ....

    Thanks.
    Last edited by fractal5; July 4th 2012 at 01:15 PM.
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  2. #2
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    Re: Solving an IVP for y(x)

    Quote Originally Posted by fractal5 View Post
    \frac{d^2y}{dx^2} = \frac{1}{a} \sqrt{1 + {(\frac{dy}{dx})^2}} ; y(0)=a>0, y'(0)=0

    let w = y'

    \frac{dw}{\sqrt{1+w^2}} = \frac{1}{a}dx

    ln|w + \sqrt{w^2 + a} | = \frac{x}{a} + c

    ln|y' + \sqrt{y'^2 + a} | = \frac{x}{a} + c

    for the given values c = 0

    y' + \sqrt{y'^2 + a} = e^\frac{x}{a}

    From here I don't know how to proceed. How do you reduce it such that I can get y(x) = ....

    Thanks.
    I would not do \displaystyle \begin{align*} \int{\frac{dw}{\sqrt{1 +w^2}}} = \ln{\left|w + \sqrt{w^2 + a}\right|} \end{align*}.

    Instead, I'd make the substitution \displaystyle \begin{align*} w = \sinh{t} \implies dw = \cosh{t}\,dt \end{align*} which would make

    \displaystyle \begin{align*} \int{\frac{dw}{\sqrt{1 + w^2}}} &= \int{\frac{\cosh{t}\,dt}{\sqrt{1 + \sinh^2{t}}}} \\ &= \int{\frac{\cosh{t}\,dt}{\sqrt{\cosh^2{t}}}} \\ &= \int{\frac{\cosh{t}\,dt}{\cosh{t}}} \\ &= \int{1\,dt} \\ &= t + C \\ &= \sinh^{-1}{w} + C \end{align*}

    So the solution to your DE becomes

    \displaystyle \begin{align*} \sinh^{-1}{w} &= \frac{x}{a} + c \\ w &= \sinh{\left(\frac{x}{a} + c\right)} \\ \frac{dy}{dx} &= \sinh{\left(\frac{x}{a} + c\right)} \end{align*}

    I'm sure you can go from here.
    Thanks from fractal5
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  3. #3
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    Re: Solving an IVP for y(x)

    Genius! Thanks a lot, I would've never thought to use the sinh function.
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