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

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.

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.