#### Explain solution RD Sharma class 12 chapter Differential Equations exercise 21.7 question 44 maths

Answer: $y=e^{\sin ^{2} x}$

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

Given: $\frac{d y}{d x}=y \sin 2 x ; y(0)=1$

Solution:

\begin{aligned} &\frac{d y}{d x}=y \sin 2 x \\\\ &\Rightarrow \frac{d y}{y}=\sin 2 x d x \end{aligned}

Integrating both sides

\begin{aligned} &\int \frac{d y}{y}=\int \sin 2 x d x \\\\ &\Rightarrow \log |y|=\frac{-\cos 2 x}{2}+c \end{aligned}                ...............(1)

Now Given that $\text { at } x=0 ; y=1[y(0)=1]$

\begin{aligned} &\log |1|=\frac{-\cos 2(0)}{2}+c \\\\ &\Rightarrow 0=\frac{-1}{2}+c \Rightarrow c=\frac{1}{2} \end{aligned}

Put in (1) we get

\begin{aligned} &\log |y|=\frac{-\cos 2 x}{2}+\frac{1}{2} \\\\ &\Rightarrow 2 \log |y|+\cos 2 x=1 \Rightarrow 2 \log y=1-\cos 2 x \\\\ &\Rightarrow \log y=\frac{2 \sin ^{2} x}{2} \end{aligned}

\begin{aligned} &{\left[1-\cos 2 x=2 \sin ^{2} x\right]} \\\\ &\Rightarrow \log y=\sin ^{2} x \\\\ &{\left[\log _{\varepsilon} a=x \Rightarrow x=e^{x}\right]} \\\\ &\Rightarrow y=e^{\sin ^{2} x} \end{aligned}