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Thread: Seperable Differential Equation

  1. #1
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    Seperable Differential Equation

    Solve the separable differential equation

    8x-3ysqrt(x^2+1)dy/dx=0

    subject to initial condition y(0) = 1

    This is what I have done so far....


    -3ysqrt(x^2+1)dy/dx = -8x
    -3ydy = -8x/sqrt(x^2+1)dx
    -3/2y^2 = ?
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by derekjonathon View Post
    Solve the separable differential equation

    8x-3ysqrt(x^2+1)dy/dx=0

    subject to initial condition y(0) = 1

    This is what I have done so far....


    -3ysqrt(x^2+1)dy/dx = -8x
    -3ydy = -8x/sqrt(x^2+1)dx
    -3/2y^2 = ?
    $\displaystyle \int3y\,dy=\int\frac{8x}{\sqrt{x^2+1}}\,dx$

    For the right integral, let $\displaystyle u=x^2+1$ so $\displaystyle du=2x\,dx$. Now it's $\displaystyle \int\frac{8}{2\sqrt{u}}\,du=8\sqrt{u}=8\sqrt{x^2+1 }$ so...
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  3. #3
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    Quote Originally Posted by derekjonathon View Post
    Solve the separable differential equation

    8x-3ysqrt(x^2+1)dy/dx=0

    subject to initial condition y(0) = 1

    This is what I have done so far....


    -3ysqrt(x^2+1)dy/dx = -8x
    -3ydy = -8x/sqrt(x^2+1)dx
    -3/2y^2 = ?
    $\displaystyle 8x - 3y\sqrt{x^2 + 1}\,\frac{dy}{dx} = 0$

    $\displaystyle 3y\sqrt{x^2 + 1}\,\frac{dy}{dx} = 8x$

    $\displaystyle y\,\frac{dy}{dx} = \frac{8}{3}\,\frac{x}{\sqrt{x^2 + 1}}$

    $\displaystyle \int{y\,\frac{dy}{dx}\,dx} = \frac{4}{3}\int{2x(x^2 + 1)^{-\frac{1}{2}}\,dx}$

    $\displaystyle \int{y\,dy} = \frac{8}{3}(x^2 + 1)^{\frac{1}{2}} + C_1$

    $\displaystyle \ln{|y|} + C_2 = \frac{8}{3}\sqrt{x^2 + 1} + C_1$

    $\displaystyle \ln{|y|} = \frac{8}{3}\sqrt{x^2 + 1} + C$ where $\displaystyle C = C_1 - C_2$

    $\displaystyle |y| = e^{\frac{8}{3}\sqrt{x^2 + 1} + C}$

    $\displaystyle |y| = e^Ce^{\frac{8}{3}\sqrt{x^2 + 1}}$

    $\displaystyle y = \pm e^Ce^{\frac{8}{3}\sqrt{x^2 + 1}}$

    $\displaystyle y = Ae^{\frac{8}{3}\sqrt{x^2 + 1}}$


    Now you know $\displaystyle y(0) = 1$


    $\displaystyle 1 = Ae^{\frac{8}{3}\sqrt{0^2 + 1}}$

    $\displaystyle 1 = Ae^{\frac{8}{3}}$

    $\displaystyle A = e^{-\frac{8}{3}}$.


    Therefore $\displaystyle y = e^{-\frac{8}{3}}e^{\frac{8}{3}\sqrt{x^2 + 1}}$

    $\displaystyle y = e^{\frac{8}{3}\sqrt{x^2 + 1} - \frac{8}{3}}$.
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  4. #4
    Super Member redsoxfan325's Avatar
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    Never mind, you fixed it.
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    Quote Originally Posted by redsoxfan325 View Post
    I think you lost a square root sign going from step 2 to step 3.
    Yes I did. Edited.
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