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#### Need Solution for R.D.Sharma Maths Class 12 Chapter 21 Differential Equations Exercise 21.8 Question 7 Maths Textbook Solution.

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

Given:$\frac{dy}{dx}=\sec \left ( x+y \right )$

Hint : -  first, we will separate variables and then solve.

Solution : -  We have,

$\frac{dy}{dx}=\sec \left ( x+y \right )$

Let $x+y=v$

Differentiating with respect to x, we get,

\begin{aligned} &\quad \frac{d}{d x}(\mathrm{x}+\mathrm{y})=\frac{d v}{d x} \\ &\Rightarrow \quad 1+\frac{d y}{d x}=\frac{d v}{d x} \\ &\Rightarrow \frac{d y}{d x}=\frac{d v}{d x}-1 \end{aligned}

Putting x + y = v and dydx  = dvdx -1 in the given differential equation, we get,

$\therefore \frac{d v}{d x}-1=\sec \mathrm{v}$

$\Rightarrow \frac{d v}{d x}-1=\frac{1}{\cos v}$

$\Rightarrow \frac{d v}{d x}=\frac{\operatorname{cos} v+1}{\cos v}$

Taking like variables on same side, we get,

$\Rightarrow \frac{\operatorname{cos} v}{\cos v+1} \mathrm{~d} \mathrm{v}=\mathrm{d} \mathrm{x} \\$

$\Rightarrow \frac{\cos v(1-\cos v)}{(1+\cos v)(1-\cos v)} d v=d x \\$

$\Rightarrow \frac{\cos v(1-\cos v)}{1-\cos ^{2} v} d v=d x \\$

$\Rightarrow \frac{\cos v(1-\cos v)}{\sin ^{2} v} d v=d x$

Integrating on both the sides, we get,

$\int \frac{\cos v(1-\cos v)}{\sin ^{2} v} d v=\int d x \\$

$\Rightarrow \int\left(\frac{\cos v}{\sin ^{2} v}-\frac{\cos ^{2} v}{\sin ^{2} v}\right) d v=\int d x \\$

$\Rightarrow \int\left(\operatorname{cotv} \operatorname{cosec} v-\cot ^{2} v\right) d v=\int d x \\$

$\Rightarrow \int\left(\operatorname{cotv} \operatorname{cosec} v d v-\left(\operatorname{cosec}^{2} v-1\right) d v=\int d x \quad\left(\because u \operatorname{sing} \cot ^{2} v=\operatorname{cosec}^{2} v-1\right)\right. \\$

$\Rightarrow-\operatorname{cosec} v+\cot v+v)=x+c$

Putting $v=x+y$,we have

$\Rightarrow-\operatorname{cosec}(x+y)+\cot (x+y)+x+y=x+c \\$

$\Rightarrow-\operatorname{cosec}(x+y)+\cot (x+y)+y=c \\$

$\Rightarrow-\frac{1}{\sin (x+y)}+\frac{\cos (x+y)}{\sin (x+y)}+y=c \\$

$\Rightarrow-\left(\frac{1-\cos (x+y)}{\sin (x+y)}\right)+y=c \\$

$\Rightarrow-\frac{2 \sin ^{2} \frac{x+y}{2}}{2 \cos \frac{x+y}{2} \sin \frac{x+y}{2}}+y=c \\$

$\Rightarrow-\tan \frac{(x+y)}{2}+y=c \\$

$\Rightarrow y=\tan \frac{(x+y)}{2}+c$

(This is the required solution)