#### Please solve RD Sharma Class 12 Chapter 21 Differential Equation Exercise Revision Exercise (RE) Question 46 maths textbook solution.

Answer : $y=\log \frac{(\operatorname{cosec} 2 x-\cot 2 x)}{y^{7}}+c$

Hint: you must know the rules of solving differential equation and integrations.

Given: $\operatorname{ysec}^{2} x+(y+7) \tan x \frac{d y}{d x}=0$

Solution : $\operatorname{ysec}^{2} x+(y+7) \tan x \frac{d y}{d x}=0$

\begin{aligned} &(y+7) \tan x \frac{d y}{d x}=-\operatorname{ysec}^{2} x \\ &(y+7) d y=\frac{-y \sec ^{2} x}{\tan x} d x \\ &\frac{(y+7)}{y} d y=-\frac{1}{\cos ^{2} x} x \frac{\cos x}{\sin x} d x \end{aligned}

\begin{aligned} &\left(1+\frac{7}{y}\right) d y=-\frac{1}{\cos x \sin x} d x \\ &\left(1+\frac{7}{y}\right) d y=-\frac{2}{2 \cos x \sin x} d x \\ &\left(1+\frac{7}{y}\right) d y=-\frac{2}{\sin 2 x} d x \end{aligned}

Now, Integrating both sides,

\begin{aligned} &\int 1 d y+\int \frac{7}{y} d y=-\frac{2}{\sin 2 x} d x \\ &y+7 \log |y|=-2 \int \operatorname{cosec} 2 x d x \\ &y+7 \log |y|=-2\left[\frac{1}{2} \log |\operatorname{cosec}(2 x)-\cot (2 x)|\right]+c \end{aligned}

\begin{aligned} &\mathrm{y}+\log \mathrm{y}^{7}=\log |\operatorname{cosec}(2 x)-\cot (2 x)|+\mathrm{c} \\ &\mathrm{y}=\log |\operatorname{cosec}(2 x)-\cot (2 x)|-\log \mathrm{y}^{7}+\mathrm{c} \\ &\mathrm{y}=\log \frac{(\operatorname{cosec} 2 \mathrm{x}-\cot 2 \mathrm{x})}{\mathrm{y}^{7}}+\mathrm{c} \end{aligned}