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Math Help - Differential Equation: Initial Value

  1. #1
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    Differential Equation: Initial Value

    \frac{dy}{dx} \sqrt{1-x^{2}} = y^{2} , y(0) = 1

    I changed the equation to get

     \frac{dy}{y^2} = \frac{dx}{\sqrt{1-x^2}}

    Took the integral to get:

     -y^{-1} = sin^{-1}(x) + C

    I'm not sure what the rules with sin^{-1}(x) are, but i got y by itself, ending with

     y = -\frac{1}{sin^{-1}(x)} + C

    When 0 is plugged in for x though, \frac{1}{sin^{-1}(0)} goes to infinity. What did i do wrong?
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    Be careful with your integration constant! It should be on the denominator once you've solved for y.
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  3. #3
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by twoteenine View Post
    \frac{dy}{dx} \sqrt{1-x^{2}} = y^{2} , y(0) = 1

    I changed the equation to get

     \frac{dy}{y^2} = \frac{dx}{\sqrt{1-x^2}}

    Took the integral to get:

     -y^{-1} = sin^{-1}(x) + C

    I'm not sure what the rules with sin^{-1}(x) are, but i got y by itself, ending with

     y = -\frac{1}{sin^{-1}(x)} + C

    When 0 is plugged in for x though, \frac{1}{sin^{-1}(0)} goes to infinity. What did i do wrong?
    Apart from your solution going to infinity, there is an error in your differentiation on the LHS of the equation you have stated

     \frac{dy}{y^2} = \frac{dx}{\sqrt{1-x^2}} gives you

     -2y^{-1} = sin^{-1}(x) + C

    or,  - sin^{-1}(x) + C'= \frac{2}{y}

    or, y = \frac{2}{- sin^{-1}(x) + C'}

    substitute y(0) = 1 to get C' = 2
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  4. #4
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    Quote Originally Posted by harish21 View Post
    Apart from your solution going to infinity, there is an error in your differentiation on the LHS of the equation you have stated

     \frac{dy}{y^2} = \frac{dx}{\sqrt{1-x^2}} gives you

     -\textcolor{red}{2}y^{-1} = sin^{-1}(x) + C

    no 2 ... twoteenine's original antideruivative is correct.

    \textcolor{red}{-\frac{1}{y} = \sin^{-1}(x) + C}

    Bruno J's suggestion follows ...

    \textcolor{red}{y = -\frac{1}{\sin^{-1}(x) + C}}

    \textcolor{red}{y(0) = 1}

    \textcolor{red}{C = -1}

    \textcolor{red}{y = \frac{1}{1 - \sin^{-1}(x)}}
    ...
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  5. #5
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by skeeter View Post
    ...
    Thanks for the correction skeeter. I did not integrate the LHS. Instead I was doing something stupid there!
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