The general solution of $\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{x+y+1}{x+y-1}$ is
- $y=x+\log (x+y)+\mathrm{c}$, where c is a constant of integration.
- $y=x-\log (x+y)+\mathrm{c}$, where c is a constant of integration.
- $y=x-\log (2 x+y)+\mathrm{c}$, where c is a constant of integration.
- $y=x^2+\log (x+y)+\mathrm{c}$, where c is a constant of integration.
Solution
Put $x+y=\mathrm{v}$...(ii) $\Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\mathrm{dv}}{\mathrm{~d} x}-1...(iii)$
Substituting (ii) and (iii) in (i), we get $\begin{aligned} & \frac{d v}{d x}-1=\frac{v+1}{v-1} \\ & \Rightarrow \frac{d v}{d x}=\frac{2 v}{v-1} \Rightarrow \frac{v-1}{2 v} d v=d x \end{aligned}$
Integrating on both sides, we get $\begin{aligned} & \frac{\mathrm{v}}{2}-\frac{1}{2} \log \mathrm{v}=x+\mathrm{c}_1 \\ & \Rightarrow \mathrm{v}-\log \mathrm{v}=2 x+2 \mathrm{c}_1 \\ & \Rightarrow x+y-\log (x+y)=2 x+2 \mathrm{c}_1 \\ & \Rightarrow y=x+\log (x+y)+\mathrm{c}, \text { where } \mathrm{c}=2 \mathrm{c}_1\end{aligned}$
Asked in: MHT CET 2024 (10 May Shift 1)