1st order Differential Equation -exact

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March 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.
Thanks in advance.
March 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.
Thanks in advance.

Read this: Exact differential - Wikipedia, the free encyclopedia

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

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