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.

Answer 1

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.

Answer 2

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

Answer 3

show work that is

Answer 4

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