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Thread: Separable Differential Equation Problem I'm having

  1. #1
    s3a
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    Separable Differential Equation Problem I'm having

    Question:
    Solve the separable differential equation
    Subject to the initial condition: , y = ?

    (My work is attached)

    Any help would be greatly appreciated!
    Thanks in advance!
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  2. #2
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    Quote Originally Posted by s3a View Post
    Question:


    Solve the separable differential equation


    Subject to the initial condition: , y = ?

    (My work is attached)

    Any help would be greatly appreciated!
    Thanks in advance!
    $\displaystyle 3x - 4y\sqrt{x^2 + 1}\,\frac{dy}{dx} = 0$

    $\displaystyle \frac{3x}{\sqrt{x^2 + 1}} - 4y\,\frac{dy}{dx} = 0$

    $\displaystyle \frac{3x}{\sqrt{x^2 + 1}} = 4y\,\frac{dy}{dx}$

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

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


    You should be able to evaluate both integrals now. To do the left hand side, you need to use the substitution $\displaystyle u = x^2 + 1$ so that $\displaystyle \frac{du}{dx} = 2x$.
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  3. #3
    s3a
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    Did you check my work? (because that's exactly what I did)

    Quote Originally Posted by Prove It View Post
    $\displaystyle 3x - 4y\sqrt{x^2 + 1}\,\frac{dy}{dx} = 0$

    $\displaystyle \frac{3x}{\sqrt{x^2 + 1}} - 4y\,\frac{dy}{dx} = 0$

    $\displaystyle \frac{3x}{\sqrt{x^2 + 1}} = 4y\,\frac{dy}{dx}$

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

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


    You should be able to evaluate both integrals now. To do the left hand side, you need to use the substitution $\displaystyle u = x^2 + 1$ so that $\displaystyle \frac{du}{dx} = 2x$.
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  4. #4
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    1. Don't be so rude - most people on this forum do not open attachments.

    2. When you have got it to the step:

    $\displaystyle \frac{3}{2}\int{u^{-\frac{1}{2}}\,du} = \int{4y\,dy}$

    $\displaystyle 3u^{\frac{1}{2}} + C_1 = 2y^2 + C_2$

    $\displaystyle 3\sqrt{x^2 + 1} + C_1 - C_2 = 2y^2$

    $\displaystyle \frac{3}{2}\sqrt{x^2 + 1} + C = y^2$, where $\displaystyle C = \frac{1}{2}(C_1 - C_2)$

    $\displaystyle y = \pm \sqrt{\frac{3}{2}\sqrt{x^2 + 1} + C}$.


    If $\displaystyle y(0) = 1$

    $\displaystyle 1 = \pm \sqrt{\frac{3}{2}\sqrt{0^2 + 1} + C}$

    $\displaystyle 1 = \frac{3}{2}\sqrt{1} + C$

    $\displaystyle 1 = \frac{3}{2} + C$

    $\displaystyle C = -\frac{1}{2}$.


    So $\displaystyle y = \pm \sqrt{\frac{3}{2}\sqrt{x^2 + 1} - \frac{1}{2}}$

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

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

    $\displaystyle y = \pm \frac{\sqrt{6\sqrt{x^2 + 1} - 2}}{2}$.
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  5. #5
    s3a
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    I get it now thanks and I didn't try to sound rude, it just probably came that way to you because it's hard to see "emotion" (for the lack of a better word) through text.
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