Question

Solve: $\frac{dy}{dx}=1-xy+y-x$

Answer 1

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

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

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

Upon Integrating both side we get,

$\ln |1+y|=x-\frac{x^2}{2}+C$

$y=-1+e^{(x-\frac{x^2}{2}+C)}$