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Thread: 1-st order DE

  1. #1
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    1-st order DE

    4xydx +(x^2+1)dy = 0
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  2. #2
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    Krizalid's Avatar
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    it's separable.
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  3. #3
    Senior Member Danneedshelp's Avatar
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    Quote Originally Posted by Link88 View Post
    4xydx +(x^2+1)dy = 0
    $\displaystyle 4xydx+(x^{2}+1)dy=0$

    $\displaystyle (x^{2}+1)dy=-4xydx$

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

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

    The left hand side becomes $\displaystyle ln(y)$.

    For the right hand side, let $\displaystyle u=x^{2}+1$, then
    $\displaystyle du=2xdx\Rightarrow\\\frac{du}{2}=xdx$

    So, $\displaystyle \int\\\frac{-4x}{x^{2}+1}=\frac{-4}{2}\int\\\frac{du}{u}$

    $\displaystyle =-2ln(u)+C=-2ln(x^{2}+1)+C$

    We are left with,

    $\displaystyle ln(y)=-2ln(x^{2}+1)+C$. Solving for y we have,

    $\displaystyle y=\frac{e^{C}}{(x^{2}+1)^{2}}$.

    you can fill in the gaps.
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  4. #4
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    $\displaystyle y=\frac{e^{C}}{(x^{2}+1)^{2}}$.

    you can fill in the gaps.
    How did you get (X^2+1)^2? : /
    Last edited by mr fantastic; Sep 28th 2009 at 09:40 PM. Reason: Added start quote tag.
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  5. #5
    Senior Member Danneedshelp's Avatar
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    Quote Originally Posted by Link88 View Post
    $\displaystyle y=\frac{e^{C}}{(x^{2}+1)^{2}}$.

    you can fill in the gaps.

    $\displaystyle e^{-2ln(x^{2}+1)+C}=e^{c}e^{-2ln(x^{2}+1)}=
    e^{c}(x^{2}+1)^{-2}=\frac{e^{c}}{(x^{2}+1)^{2}}$

    Remember your log properties. $\displaystyle log(x^{c})=clog(x)$. So, in this case, e cancled out the natural log and the -2 out front jumped back up to its exponent position.
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