The general solution of the differential equation $\frac{d y}{d x}=\frac{x+y+1}{x+y-1}$ is given by

The general solution of the differential equation $\frac{d y}{d x}=\frac{x+y+1}{x+y-1}$ is given by
  1. $y=x \log (x+y)+c$
  2. $x-y=\log (x+y)+c$
  3. $x+y=\log (x+y)+c$
  4. $y=x+\log (x+y)+c$

Solution

$\frac{d y}{d x}=\frac{x+y+1}{x+y-1}$ Put $x+y=u \Rightarrow 1+\frac{d y}{d x}=\frac{d u}{d x}$ $\begin{aligned} & \therefore \frac{\mathrm{du}}{\mathrm{dx}}-1=\frac{\mathrm{u}+1}{\mathrm{u}-1} \Rightarrow \frac{\mathrm{du}}{\mathrm{dx}}=\frac{\mathrm{du}+1}{\mathrm{dx}-1}+1=\frac{2 \mathrm{u}}{\mathrm{u}-1} \\ & \therefore\left(\frac{\mathrm{u}-1}{\mathrm{u}}\right) \mathrm{du}=2 \mathrm{dx} \end{aligned}$ Integrating both sides, we get $\begin{aligned} & \int d u-\int \frac{d u}{u}=\int 2 d x \\ & \therefore u-\log u=2 x+c \\ & \therefore x+y-\log (x+y)=2 x+c \Rightarrow y=x+\log (x+y)+c \end{aligned}$

Asked in: MHT CET 2021 (20 Sep Shift 1)

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