The general solution of the differential equation e^x dy + (y e^x + 2x) dx = 0 is

It is given that e^xdy + (ye^x + 2x) dx = 0

This is equation in the form of (where, p = 1 and Q = -2xe^-x)

Now, I.F. =

Thus, the solution of the given differential equation is given by the relation:

y(I.F.) =

⇒ ye^x = -x² + C

⇒ ye^x + x² = C