Question

E^x+ydx + e^2y-xdy = 0?

`e^(x+y)dx + e^(2y-x)dy = 0?`

Answer 1

`1/2 e^(2x)+e^y=C`

Explanation:

The equation

`e^(x+y)dx + e^(2y-x)dy = 0`

is easy to rewrite in the form

`e^(2x)dx+e^ydy=0`

Integration yields

`1/2 e^(2x)+e^y=C`

where `C` is a constant of integration.

Answer 2

`"The GS : "e^(2x)+2e^y=C,`

Explanation:

I hope, the Question is to find the General Solution (GS) of the

Diff. Eqn. (DE) : `e^(x+y)dx+e^(2y-x)dy=0`.

Multiplying the given eqn. by `e^(x-y)`, we get,

`e^{(x+y)+(x-y)}dx+e^{(2y-x)+(x-y)}dy=0, or,`

`e^(2x)dx+e^ydy=0`.

This is a Separable Variable Type DE.

To have its GS, integrating, `inte^(2x)dx+inte^ydy=c`.

`:. 1/2e^(2x)+e^y=c, or,`

`e^(2x)+2e^y=C,` is the desired GS, where, `C=2c`.

Enjoy Maths.!