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Math Help - find implicit form

  1. #1
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    find implicit form

    hi
    question reads
    find in implicit form the general solution of differential equation


    \frac{dy}{dx}=3y^2e^{-2x}\sqrt{8+e^{-2x}}

    = \int{3y^2}=\int{e^{-2x}}\sqrt{8+e^{-2x}}

    = \int{3y^2}=\int{e^{-2x}}+e^{\frac{-3}{2}}+8^{\frac{1}{2}}

    y^3=\frac{-1}{2}e^{-2x}-\frac{2}{3}e^{\frac{3}{2}}+\frac{2}{3}8x^\frac{3}{  2}

    i did try but as you can see not sure what i am doing could do with some expert help please
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by smartcar29 View Post
    hi
    question reads
    find in implicit form the general solution of differential equation


    \frac{dy}{dx}=3y^2e^{-2x}\sqrt{8+e^{-2x}}

    = \int{3y^2}=\int{e^{-2x}}\sqrt{8+e^{-2x}}

    = \int{3y^2}=\int{e^{-2x}}+e^{\frac{-3}{2}}+8^{\frac{1}{2}}

    y^3=\frac{-1}{2}e^{-2x}-\frac{2}{3}e^{\frac{3}{2}}+\frac{2}{3}8x^\frac{3}{  2}.

    i did try but as you can see not sure what i am doing could do with some expert help please
    When you separate the variables, you get \frac{\,dy}{y^2}=3e^{-2x}\sqrt{8+e^{-2x}}\,dx. So when you integrate the LHS, you get -\frac{1}{y}+C. When you integrate the RHS, you apply the substitution u=8+e^{-2x}\implies \,du=-2e^{-2x}\,dx. So it follows that 3\int e^{-2x}\sqrt{8+e^{-2x}}\,dx=-\tfrac{3}{2}\int\sqrt{u}\,du=-u^{3/2}+C=-\left(8+e^{-2x}\right)^{3/2}+C

    So it follows now that \frac{1}{y}+C=-\left(8+e^{-2x}\right)^{3/2}+C\implies \frac{1}{y}=-\left(8+e^{-2x}\right)^{3/2}+K. This would be the implicit solution.

    The explicit solution would be y=\frac{1}{K-\left(8+e^{-2x}\right)^{3/2}}
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  3. #3
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    thank you for your much needed help i really suck at math, going to spend less time playing xbox and more time doing revision.

    thanks again.
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