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Thread: Solve y'+y^2sinx=0.

  1. April 10th 2011, 04:04 PM #1
    Taurus3
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    Solve y'+y^2sinx=0.

    solve the differential equation:
    y'+y^2sinx=0

    I used the method of separation:

    y'=-y^2sinx
    dy/dx=-y^2sinx
    dy/-y^2=sinxdx
    integral of -y^-2dy=integral of sinxdx
    1/y=cosx+C
    y=1/(cosx+C)

    Did I do this right?
    Last edited by mr fantastic; April 10th 2011 at 08:17 PM. Reason: Re-titled.
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  2. April 10th 2011, 04:08 PM #2
    dwsmith
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    Quote Originally Posted by Taurus3 View Post
    solve the differential equation:
    y'+y^2sinx=0

    I used the method of separation:

    y'=-y^2sinx
    dy/dx=-y^2sinx
    dy/-y^2=sinxdx
    integral of -y^-2dy=integral of sinxdx
    1/y=cosx+C
    y=1/(cosx+C)

    Did I do this right?
    \displaystyle\frac{y'}{y^2}=-\sin(x)\Rightarrow \frac{-1}{y}=\cos(x)+C\Rightarrow y=\frac{-1}{\cos(x)+C}
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