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#### Explain Solution R.D.Sharma Class 12 Chapter 21 Differential Equations Exercise Revision Exercise Question 36 Maths Textbook Solution.

Answer: $y=\tan \left ( \frac{x+y}{2} \right )+C$

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

Given:$\cos \left ( x+y \right )dy=dx$

Solution:$\cos \left ( x+y \right )dy=dx$                        $\left ( \frac{dy}{dx}=\frac{1}{\cos \left ( x+y \right )} \right )$

Let $\left ( x+y \right )=u$ and differentiating both sides,

\begin{aligned} &1+\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{du}}{\mathrm{dx}} \\ &\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{du}}{\mathrm{dx}}-1 \\ &\frac{1}{\cos \mathrm{u}}=\left(\frac{\mathrm{du}}{\mathrm{dx}}-1\right) \\ &1=\left(\frac{\mathrm{du}}{\mathrm{dx}}-1\right) \cdot \cos \mathrm{u} \\ &\operatorname{Cos} \mathrm{u} \frac{\mathrm{du}}{\mathrm{dx}}=1+\cos \mathrm{u} \end{aligned}

\begin{aligned} &\frac{\text { cosu }}{1+\operatorname{cosu}} d u=d x \\ &\frac{(1+\cos u)-1}{1+\cos u} d u=d x \\ &{\left[\frac{1+\cos u}{1+\operatorname{cosu}}-\frac{1}{1+\operatorname{cosu}}\right] d u=d x} \\ &{\left[1-\frac{1}{1+\operatorname{cosu}}\right] d u=d x} \end{aligned}

\begin{aligned} &\Rightarrow\left[1-\frac{1}{2 \cos ^{2} \frac{u}{2}}\right] d u=d x \quad\left[\because \frac{1}{1+\cos x}=2 \cos ^{2} \frac{u}{2}\right. \\ &\Rightarrow\left[1-\frac{1}{2} \sec ^{2} \frac{u}{2}\right] d u=d x \end{aligned}

Now, integrating both sides,

\begin{aligned} &\int 1 \mathrm{du}=\int 1 .-\frac{1}{2} \sec ^{2} \frac{u}{2} \mathrm{du}=\int \mathrm{dx} \\ &\mathrm{U}-\tan \frac{\mathrm{u}}{2}=\mathrm{x}+\mathrm{c} \quad\left[\because \int \sec ^{2} \frac{\mathrm{x}}{2}=2 \tan \frac{\mathrm{x}}{2}\right] \end{aligned}

Put the value of u

\begin{aligned} &(x+y)-\tan \frac{(x+y)}{2}=x+c \quad[\because u=x+y] \\ &\therefore y=\tan \left(\frac{x+y}{2}\right)+C \end{aligned}