Question

Find the general solution of the differential equation 

ye^x/y dx = (xe^x/y + y²) dy, y ≠ 0

Answer 1

Given ye^x/y dx = (xe^x/y + y²) dy, y ≠ 0

⇒ `dy/dx = (ye^(x//y))/(xe^(x//y) + y^2)`

⇒ `dy/dx = (xe^(x//y) + y^2)/(ye^(x//y))`

Put x = vy

`dx/dy = v + y (dv)/dy`

⇒ `v + y (dv)/dy = ("vye^v + y^2)/(ye^v)`

⇒ `y (dv)/dy = (vye^v + y^2)/(ye^v) - v`

⇒ `y (dv)/dy = (vye^v + y^2 - vye^v)/(ye^v)`

⇒ `y (dv)/dy = y^2/(ye^v)`

⇒ `(dv)/dy = 1/e^v`

⇒ `int e^v dv = int dy + "C"`

⇒ e^v = y + C

⇒ e^x/y = y + C is the required solution.