Question

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

Answer 1

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

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

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

$\ln (1+y)=\log x+x+c$

$\Rightarrow \ln (1+y)-\log x=x+c$

$\Rightarrow \frac{1+y}{x}=e^{x+c}$

$\Rightarrow 1+y=x{e}^{x+c}$

$\Rightarrow y=x{e}^{x+c}-1$