Question

Find the general solution of the differential equation `dy/dx + 1/x = e^y/x`.

Answer 1

`dy/dx + 1/x = e^y/x`

`dy/dx = (e^y - 1)/x`

`1/(e^y - 1) dy = 1/x dx`

Integrating both side

`int 1/(e^y - 1) dy = int 1/x dx`

`int 1/(e^y(1 - e^-y)) dy = int 1/x dx`

`int (e^-y)/(1 - e^-y) dy = int 1/x dx`

Let 1 – e^–y = t

0 – e^–y (–1) dy = dt

e^–y dy = dt

`int 1/t dt = int 1/x dx`

log t = log x + log C

log (1 – e^–y) = log xC

1 – e^–y = xC

`1 - xC = 1/e^y`

 `e^y = 1/(1 - xC)`

`y = e^((Cx - 1))`