Question

$\frac{dy}{dx}=\frac{xy+y}{xy+x}$ ,then the solution of differential equation is

• A. $y=x{e}^x+c$
• B. $y=e^x+c$
• C. $y=x{e}^{x-y}+c$
• D. $y=x+c$

Answer 1

The correct option is C $y=x{e}^{x-y}+c$

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

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

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

Now, integrate on both sides we get

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

$\log \frac{y}{x}=x-y+\log C$

$\frac{y}{x}=c{e}^{x-y}$

$y=xc{e}^{x-y}$