#### Provide solution for RD Sharma maths class 12 chapter Differential Equations exercise 21.7 question 42

Answer: $4 \cdot y^{2}\left(2-e^{x}\right)=1$

Hint: Separate the terms of x and y and then integrate them.

Given: $\frac{d y}{d x}=2 e^{x} y^{3}, y(0)=\frac{1}{2}$

Solution:

\begin{aligned} &\frac{d y}{d x}=2 e^{x} y^{3} \\\\ &\frac{d y}{y^{3}}=2 e^{x} d x \end{aligned}

Integrating both sides

\begin{aligned} &\int \frac{d y}{y^{3}}=2 \int e^{x} d x \\\\ &\Rightarrow \int y^{-3} d y=2 \int e^{x} d x \Rightarrow \frac{y^{-3+1}}{-3+1}=2 e^{x}+c \\\\ &{\left[\int x^{n} d x=\frac{x^{n+1}}{n+1}\right]} \end{aligned}

\begin{aligned} &\Rightarrow \frac{y^{-2}}{-2}=2 e^{x}+c \\\\ &\Rightarrow \frac{-1}{2 y^{2}}=2 e^{x}+c \\\\ &\Rightarrow-1=2 y^{2}\left(2 e^{x}+c\right) \end{aligned}                        ..............(1)

Now given that

$y(0)=\frac{1}{2} \text { i.e. at } x=0\: \&\; y=\frac{1}{2}$

Put in (1)

\begin{aligned} &-1=2\left(\frac{1}{2}\right)^{2}\left(2 e^{0}+c\right) \\\\ &\Rightarrow-1=\left(\frac{1}{2}\right)(2(1)+c) \Rightarrow-2=2+c \Rightarrow c=-4 \end{aligned}

Put in (1) we get

\begin{aligned} &\Rightarrow-1=2 y^{2}\left(2 e^{x}-4\right) \Rightarrow-1=-2\left(2 y^{2}\right)\left(2-e^{x}\right) \\\\ &\Rightarrow 4 y^{2}\left(2-e^{x}\right)=1 \end{aligned}