Hi,

I would like to know how do I show that:

y-(e^x) cos y + [x+(e^x) sin y]y' = 0 is an exact differential equation.

Thanks in advance.

Printable View

- Mar 29th 2009, 12:02 AMzxcv1st order Differential Equation -exact
Hi,

I would like to know how do I show that:

y-(e^x) cos y + [x+(e^x) sin y]y' = 0 is an exact differential equation.

Thanks in advance. - Mar 29th 2009, 01:47 AMmr fantastic
Read this: Exact differential - Wikipedia, the free encyclopedia

You have $\displaystyle (y - e^x \cos y) \, dx + (x + e^x \sin y) \, dy = 0$

where $\displaystyle A = y - e^x \cos y$ and $\displaystyle B = x + e^x \sin y$.

Show $\displaystyle \frac{\partial A}{\partial y} = \frac{\partial B}{\partial x}$.