Question

Solve:
$(1-y)x\frac{dy}{dx}+(1+x)y=0$

Answer 1

Consider the given differential equation,

$(1-y)x\frac{dy}{dx}+(1+x)y=0$

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

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

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

Taking integration both sides with respect to x, we get

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

$\log y-y=-\log x-x+C$

$\log y+\log x+=x-y+C$

$\log x.y=x-y+C$

$xy=e^{x-y+C}$

Hence this is the answer.