The general solution of the differential equation $\frac{\mathrm{d} y}{\mathrm{~d} x}=1-x+y-x y$ is (where…

The general solution of the differential equation $\frac{\mathrm{d} y}{\mathrm{~d} x}=1-x+y-x y$ is (where $C$ is a constant of integration)
  1. $\log (1+y)=x+\frac{x^2}{2}+C$
  2. $\log (1-x)=\log (1+y)+y+C$
  3. $\log (1+y)=y-\frac{x^2}{2}+C$
  4. $\log (1+y)=x-\frac{x^2}{2}+C$

Solution

$\begin{aligned} & \frac{\mathrm{d} y}{\mathrm{~d} x}=1-x+y-x y=(1-x)(1+y) \\ & \Rightarrow \int \frac{\mathrm{d} y}{1+y}=\int(1-x) \mathrm{d} x \\ & \Rightarrow \log (1+y)=x-\frac{x^2}{2}+C\end{aligned}$

Asked in: MHT CET 2022 (08 Aug Shift 1)

Practice more Differential Equations questions on Aicharya