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Math Help - 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 = ?
    \int3y\,dy=\int\frac{8x}{\sqrt{x^2+1}}\,dx

    For the right integral, let u=x^2+1 so du=2x\,dx. Now it's \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 = ?
    8x - 3y\sqrt{x^2 + 1}\,\frac{dy}{dx} = 0

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

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

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

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

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

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

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

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

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

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


    Now you know y(0) = 1


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

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

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


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

    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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  5. #5
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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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