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Math Help - Seperable Intial-Value Problem

  1. #1
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    Seperable Intial-Value Problem

     (x^2+3y^2)dx + (2xy)dy = 0
    y(2)=6

    so this homogeneous eqn. needs to be transformed into a separable eqn. right?
    y=vx , v=\frac{y}{x}


     \frac{dy}{dx}=\frac{x^2+3y^2}{2xy} or is it \frac{x^2+3y^2}{2x}dx - \frac{y}{1}dy= How can I plug in \frac{y}{x} here ?

    \frac {dy}{dx}= v + x\frac{dv}{dx}

    Not sure how to solve this one.
    Last edited by JJ007; January 30th 2011 at 09:53 AM.
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  2. #2
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    Quote Originally Posted by JJ007 View Post
     (x^2+3y^2)dx + (2xy)dy = 0
    y(2)=6

    so this homogeneous eqn. needs to be transformed into a separable eqn. right?
    y=vx , v=\frac{y}{x}


     \frac{dy}{dx}=\frac{x^2+3y^2}{2xy}= How can I plug in \frac{y}{x} here ?

    \frac {dy}{dx}= v + x\frac{dv}{dx}

    Not sure how to solve this one.
    Multiply the numerator and denominator by

    \frac{1}{x^2}

    This gives

    \displaystyle \frac{1+3\left( \frac{y}{x}\right)^2}{2\left( \frac{y}{x}\right)}
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  3. #3
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    Quote Originally Posted by JJ007 View Post
    \frac {dy}{dx}= v + x\frac{dv}{dx}
    You should write the above as follows:

    y=vx\Rightarrow dy=xdv+vdx

    You would then plug this into dy of the DE.
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