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Thread: Differential equation problem

  1. #1
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    Differential equation problem

    Hello,

    I'm trying to find the integration factor $\displaystyle \lambda(x,y)$ of this differential equation: $\displaystyle (siny-3x^2cosy)cosydx+xdy=0$, with exception that $\displaystyle \lambda$ being a function, dependent in only one variable, x or y.

    Any ideas on solving this equation?

    Thanks
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: Differential equation problem

    Quote Originally Posted by patzer View Post
    Hello,

    I'm trying to find the integration factor $\displaystyle \lambda(x,y)$ of this differential equation: $\displaystyle (siny-3x^2cosy)cosydx+xdy=0$, with exception that $\displaystyle \lambda$ being a function, dependent in only one variable, x or y.

    Any ideas on solving this equation?

    Thanks
    Hint: If $\displaystyle Pdx+Qdy=0$ and $\displaystyle \lambda(x,y)=\mu (z)$, the equality $\displaystyle (\mu P)_y=(\mu Q)_x$ sometimes (only sometimes) allows to predict the form of $\displaystyle z$. Take into account that in general, if we don't know a priori the form of the integration factor, there is no a general method to find it.
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  3. #3
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    Re: Differential equation problem

    Hello sir,

    Thank you for your help.

    This is my progress so far:

    I know that $\displaystyle \frac{d\lambda}{\lambda}=-\frac{\frac{\partial P}{\partial y}-\frac{\partial Q}{\partial x}}{P}dy$ where $\displaystyle \frac{\frac{\partial P}{\partial y}-\frac{\partial Q}{\partial x}}{P}$=$\displaystyle \frac{3x^2sin(2y)+cos(2y)-1}{3x^2(cosy)^2-cosysiny}$

    My problem now is with this expression $\displaystyle \frac{3x^2sin(2y)+cos(2y)-1}{3x^2(cosy)^2-cosysiny}$, I simply don't know how to simplify it.
    Wolfram Alpha gives $\displaystyle 2tan(y)$ as the answer. Can you help me on this point?

    Thank you again.
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