this is linear algebra & differential equation problem level. I need to find the exact equation of
ex: (e^2y-xe^y)y' - e^y-x = 0 and at the end i need to solve for y.

Question

this is linear algebra & differential equation problem level. I need to find the exact equation of 
ex: (e^2y-xe^y)y' - e^y-x = 0 and at the end i need to solve for y.

anonymous · Answer

You've got an exact DE.
$$(e^{2y}-xe^y)dy+(-e^y-x)dx=0$$Let $M=-e^y-x$ and $N=e^{2y}-xe^y$.$$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$$There exists some $f(x, y)=c$ for which$$\frac{\partial f}{\partial x}=M$$and$$\frac{\partial f}{\partial y}=N$$since$$f_{xy}=f_{yx}$$Therefore, $$f(x, y)=c=\int Mdx\cup\int Ndy$$I'm using the term "union" loosely here.  The point is that you take distinct terms from each integral.

adunb8 · Answer

? i dont quite get it.. can you the steps?

adunb8 · Answer

show work that is

anonymous · Answer

http://www.wolframalpha.com/input/?i=%28e^%282y%29-xe^y%29y%27+-+e^y-x+%3D+0