Differential equation problem

• Jan 30th 2013, 10:21 AM
patzer
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
• Jan 30th 2013, 12:17 PM
FernandoRevilla
Re: Differential equation problem
Quote:

Originally Posted by patzer
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.
• Jan 30th 2013, 03:48 PM
patzer
Re: Differential equation problem
Hello sir,

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?