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#### Please Solve RD Sharma Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 25 Maths Textbook Solution.

Answer:  $x=\tan y+C \sqrt{\tan y}$

Hint: To solve this equation we use $\tan x$  and convert it to  $\sin x$  and  $\cos x$

Give:  \begin{aligned} &(x+\tan y) d y=\sin 2 y d x \\ & \end{aligned}

Solution: $(x+\tan y) d y=\sin 2 y d x$

\begin{aligned} &=\frac{d x}{d y}=\frac{(x+\tan y)}{\sin 2 y} \\ &=\frac{d x}{d y}=\frac{x}{\sin 2 y}+\frac{\tan y}{\sin 2 y} \\ &=\frac{d x}{d y}=x \operatorname{cosec} 2 y+\frac{\frac{\sin y}{\cos y}}{2 \sin y \cos y} \\ &=\frac{d x}{d y}=x \operatorname{cosec} 2 y+\frac{\frac{\sin y}{\cos y}}{2 \sin y \cos y} \end{aligned}

\begin{aligned} &=\frac{d x}{d y}=x \operatorname{cosec} 2 y+\frac{1}{2 \cos ^{2} y} \\ &=\frac{d x}{d y}=x \operatorname{cosec} 2 y+\frac{1}{2} \sec ^{2} y \\ &=\frac{d x}{d y}+R x=S \end{aligned}

\begin{aligned} &R=\operatorname{cosec} 2 y, S=\frac{1}{2} \sec ^{2} y \\ & \end{aligned}

$\text { If }=e^{\int R d y} \\$

$=e^{\int(-\operatorname{cosec} 2 y) d y} \\$

$=e^{-\log |\operatorname{cosec} 2 y-\cot 2 y|} \\$

$=\operatorname{cosec} 2 y-\cot 2 y \\$

$=\frac{1}{\sin 2 y}-\frac{\cos 2 y}{\sin 2 y} \\$

$=\frac{1-\cos 2 y}{\sin 2 y}$

\begin{aligned} &=\frac{2 \sin ^{2} y}{2 \sin y \cos y} \\ & \end{aligned}

$=\frac{\sin y}{\cos y} \\$

$=\tan y \\$

$=e^{-\log |\tan y|} \\$

$=e^{\log |\cot y|} \\$

$=\cot y$

\begin{aligned} &=x I f=\int S I f+C \\ &=x \cot y=\int \frac{1}{2} \sec ^{2} y \cot y d y+C \\ &=x \cot y=\int \frac{1}{2 \cos ^{2} y} \frac{\cos y}{\sin y} d y+C \\ &=x \cot y=\int \frac{1}{2 \cos y} \frac{1}{\sin y} d y+C \\ &=x \cot y=\int \frac{1}{\sin 2 y} d y+C \\ &=x \cot y=\frac{1}{2} \log |\operatorname{cosec} 2 y-\cot 2 y|+C \end{aligned}