The particular solution of the differential equation $\frac{d y}{d x}-e^x=y e^x$, when $x=0$ and $y=1$ is

The particular solution of the differential equation $\frac{d y}{d x}-e^x=y e^x$, when $x=0$ and $y=1$ is
  1. $\log \left(\frac{y+1}{2}\right)=\frac{e^x}{2}-\frac{1}{2}$
  2. $\log \left(\frac{y+1}{2}\right)=e^x-1$
  3. $\log (y-1)=e^x-1$
  4. $\log 2(y-1)=e^x-1$

Solution

$\begin{aligned} & \frac{d y}{d x}-e^x=y e^x \Rightarrow \frac{d y}{d x}=(y+1) e^x \Rightarrow \int \frac{d y}{y+1}=\int e^x d x \\ & \Rightarrow \log (y+1)=e^x+c \\ & \text { for } x=0, y=1 \Rightarrow c=\log 2-1 \\ & \text { Hence, } \log (y+1)=e^x+\log 2-1 \\ & \Rightarrow \log (y+1)-\log 2=e^x-1 \\ & \Rightarrow \log \left(\frac{y+1}{2}\right)=e^x-1\end{aligned}$

Asked in: MHT CET 2022 (07 Aug Shift 1)

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