Answer :

Solution :

(1 + y)$\frac{dy}{dx}$ = y

$\frac{(1+y)}{y}$dy = dx

$\frac{1}{y}$dy + 1dy = dx

Integrating both side,

$\int \frac{1}{y}dy+\int 1dy=\int dx+C$

lny + y = x + C

at x = 1, y = 1

ln1 + 1 = 1 + C

C = 0

So, lny = (x - y)

y = $e^{\left(x-y\right)}$ = $\frac{e^x}{e^y}$

ye^y = e^x