Question

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

Answer 1

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

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

Put $x+y=t$........(ii)
Differentiating with respect to $y$, we get

$\Rightarrow \frac{dx}{dy}+1=\frac{dt}{dy}$
$\Rightarrow \frac{dx}{dy}=\frac{dt}{dy}-1$.....(iii)
Substitute equations (ii) and (iii) in equation(i), we get

$\Rightarrow (t+1)=\frac{dt}{dy}-1$
$\Rightarrow t+1+1=\frac{dt}{dy}$

$\Rightarrow dy=\frac{1}{t+2}dt$
Integrating, we get
$\Rightarrow \int dy=\int \frac{1}{t+2}dt$
$\Rightarrow y=\log |t+2|$ [Since, $\int \frac{f^{\prime }\left(x\right)}{f\left(x\right)}=\log f\left(x\right)+c$]

$\therefore y=\log |x+y+2|$ [From equation(ii)]