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