Question

The solution of $(x+y+1)\frac{dy}{dx}=1$ is:

• A. $x=-(y+2)+c{e}^y$
• B. $y=-(x+2)+c{e}^x$
• C. $x=-(y+2)+c{e}^x$
• D. $x=(y+2)+c{e}^{-y}$

Answer 1

Answer: (A)

$x=-(y+2)+c{e}^y$

Let $x+y+1=v$
$\Rightarrow \frac{dv}{dx}=1+\frac{dy}{dx}$
$\Rightarrow \frac{dy}{dx}=(\frac{dv}{dx}-1)$
$\Rightarrow (\frac{dv}{dx}-1)=\frac{1}{v}$
$\frac{dv}{dx}=\frac{1+v}{v}$
$\left(\frac{v}{1+v}\right)dv=dx$
$\Rightarrow \int (1-\frac{1}{1+v})dv=\int dx+c_1$
$\Rightarrow v-log(1+v)=x+c_1$
$(y+1)-log(x+y+2)=c_1$
$\Rightarrow c{e}^{y+1}=x+y+2$
$\Rightarrow c{e}^y=x+y+2$
$x=-(y+2)+c_1{e}^y$