Here, putting x = kx and y = ky

= k⁰f(x,y)

Therefore, the given differential equation is homogeneous.

To solve it we make the substitution.

x = vy

Differentiation eq. with respect to x, we get

Integrating both sides, we get

Put e^v + v = t

(e^v + 1)dv = dt

logt

log(e^v + v)

∴ log(e^v + v) = - logy + logC (∴ From (i) eq.)

Multiply by y on both side, we get

ye^x/y + x = C

x + ye^x/y = C

The required solution of the differential equation.