#### Please Solve R.D. Sharma class 12 Chapter 21  Differential Equations Exercise 21.8 Question 1 Maths textbook Solution.

Answer: $\tan ^{-1}\left ( x+y+1 \right )=x+c$

Given:$\frac{dy}{dx}=\left ( x+y+1 \right )^{2}$

To find    :-  solve the differential equation.

Hint          :-  differential equation of the form

$\frac{dy}{dx}=\left ( ax+by+c \right )$can be reduced to variable separable form by the substitution $ax+by+c=v$

Solution:-we have

$\frac{dy}{dx}=\left ( x+y+1 \right )^{2}$                            ........(i)

$Let\: x+y+1=v$

Differentiating with respect to x, we get

\begin{aligned} &\Rightarrow 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}                                                                    ....(ii)

Now, substituting equation (ii) in equation (i), we get,

\begin{aligned} & \frac{d v}{d x}-1=(x+y+1)^{2} \\ \Rightarrow & \frac{d v}{d x}-1=v^{2} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad(\because x+y+1=v) \\ \Rightarrow & \frac{d v}{d x}=v^{2}+1 \\ \Rightarrow & \frac{d v}{v^{2}+1}=d x \end{aligned}( taking like variable on same side)

Integrating on both sides, we get,

$\int \frac{dv}{v^{2}+1}=\int dx$

$\Rightarrow \tan ^{-1}v=x+c$                                                                $\left ( \because \int\frac{dx}{a^{2+}x^{2}}=\frac{1}{a}\tan ^{-1}\left ( \frac{x}{a} \right )\right )$

Putting the value of V, we get,

$\Rightarrow \tan ^{-1}\left ( x+y+1 \right )=x+c$            ,                    which is the required solution.