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

It is given that e^{x}dy + (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} = -x^{2} + C

⇒ ye^{x} + x^{2} = C

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