Let F(x,y) be the curve passing through origin and let (x,y) be a point on the curve.

We know the slope of the tangent to the curve at (x,y) is .

According to the given conditions, we get,

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

Now, I.F. =

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

y(I.F.) =

--------(1)

Now,

Thus, from equation (1), we get,

⇒ y = -(x+1) + Ce^x

⇒ x + y + 1 = Ce^x -----(2)

Now, it is given that curve passes through origin.

Thus, equation (2) becomes:

1 = C

⇒ C = 1

Substituting C = 1 in equation (2), we get,

x + y – 1 = e^x

Therefore, the required general solution of the given differential equation is

x + y -1 = e^x