# 1st order Differential Equation -exact

• Mar 29th 2009, 12:02 AM
zxcv
1st 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.
• Mar 29th 2009, 01:47 AM
mr fantastic
Quote:

Originally Posted by zxcv
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.
You have $(y - e^x \cos y) \, dx + (x + e^x \sin y) \, dy = 0$
where $A = y - e^x \cos y$ and $B = x + e^x \sin y$.
Show $\frac{\partial A}{\partial y} = \frac{\partial B}{\partial x}$.