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Math Help - a differential equation

  1. #1
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    a differential equation

    would anyone be able to show us how you would find the general solution of this differential equation in implicit form for

    \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2y^{1/2}(2e^{2x}-5)}{3(e^{2x}-5x)^{2/3}} (y>0)
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  2. #2
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    Re: a differential equation

    Pretty close to trivial, isn't it? That's clearly a separable equation:
    y^{-1/2}dy= \frac{2}{3}\frac{\left(e^{2x}- 5x\right)^{-2/3}}{2e^{2x}- 5}dx

    Integrate. (Do you recognize that (e^{2x}- 5x)'= 2e^{2x}- 5?)
    Last edited by HallsofIvy; March 16th 2012 at 06:37 AM.
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  3. #3
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    Re: a differential equation

    yes I see that. I have only started calculus a few weeks ago so I do apologize. It is the y and what is ment by implicit form that is throwing me
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  4. #4
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    Re: a differential equation

    "Implicit form" simply means that you don't have to solve for y. Clearly when you integrate the left side you have something involving y^{1/2}. Leave it in that form.
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  5. #5
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    Re: a differential equation

    So would the correct answer be

    \frac{y^{1/2}}{1/2}=6(2e^{2x}-5x)^{1/3}+c
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  6. #6
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    Re: a differential equation

    Quote Originally Posted by Orlando View Post
    would anyone be able to show us how you would find the general solution of this differential equation in implicit form for

    \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2y^{1/2}(2e^{2x}-5)}{3(e^{2x}-5x)^{2/3}} (y>0)
    \int \frac{dy}{\sqrt y} = \int \frac{2(2e^{2x}-5)}{3(e^{2x}-5x)^{2/3}}\,dx

    Hence :

    2\sqrt y=2\sqrt[3]{e^{2x}-5x} + C \Rightarrow \sqrt y=\sqrt[3]{e^{2x}-5x} + C
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