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Thread: Implicit form question

  1. #1
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    Question Implicit form question

    Hi, I am able to find the general solution in implicit form for single line equations but I'm really having trouble with this & would appreciate some assistance.

    $\displaystyle \frac{dy}{dx}=\frac{6y^\frac{2}{3} sin x}{(2 - cos x)^3}$

    Thanks
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  2. #2
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    Did you try separating variables?.

    $\displaystyle \frac{1}{6}\int y^{\frac{-2}{3}}dy=\int\frac{sin(x)}{(2-cos(x))^{3}}dx$

    Integrate both sides and don't forget the constant on the right.
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  3. #3
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    Quote Originally Posted by galactus View Post
    Did you try separating variables?.

    $\displaystyle \frac{1}{6}\int y^{\frac{-2}{3}}dy=\int\frac{sin(x)}{(2-cos(x))^{3}}dx$

    Integrate both sides and don't forget the constant on the right.
    Thank you & I'll give it a go now.
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  4. #4
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    Okay so integrating both sides gives me this:

    $\displaystyle \frac{y^5/3}{10}=-\frac{1}{2(2-cos x)^2} + c$

    Is this the implicit form or do I need to go further?

    I have an initial condition $\displaystyle y(0)=1$ which I presume I can solve for c & then substitute back into this equation?
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  5. #5
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    Yes, use your initial condition

    $\displaystyle \frac{1}{2}y^{\frac{1}{3}}=\frac{-1}{2(cos(x)-2)^{2}}+C$

    $\displaystyle y^{\frac{1}{3}}=\frac{-1}{(cos(x)-2)^{2}}+2C$

    $\displaystyle y=\left(\frac{-1}{(cos(x)-2)^{2}}+C_{1}\right)^{3}$
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  6. #6
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    Quote Originally Posted by galactus View Post
    Yes, use your initial condition

    $\displaystyle \frac{1}{2}y^{\frac{1}{3}}=\frac{-1}{2(cos(x)-2)^{2}}+C$

    $\displaystyle y^{\frac{1}{3}}=\frac{-1}{(cos(x)-2)^{2}}+2C$

    $\displaystyle y=\left(\frac{-1}{(cos(x)-2)^{2}}+C_{1}\right)^{3}$
    Right I can see how you've done that (including the initial fraction for y which I left undone!) but where does the $\displaystyle +C_{1}$ come from?
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  7. #7
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    Quote Originally Posted by bigroo View Post
    Right I can see how you've done that (including the initial fraction for y which I left undone!) but where does the $\displaystyle +C_{1}$ come from?
    I'm still not sure about that $\displaystyle +C_{1}$ so I've left it as $\displaystyle +2C$. Maybe someone will tell me otherwise. Also why did the denominator reverse?

    Substituting back into general equation for $\displaystyle y(0)=1$:

    $\displaystyle 1=\left(\frac{-1}{(2-cos(0))^{2}}+2C\right)^{3}$

    $\displaystyle C=1$

    Hence $\displaystyle y=\left(\frac{-1}{(cos(x)-2)^{2}}+2\right)^{3}$

    Would this satisfy my original question now or have I made it explicit? Thanks
    Last edited by bigroo; Dec 9th 2009 at 07:29 AM.
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