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Please Solve R.D.Sharma class 12 Chapter 21  Differential Equations Exercise 21.8 Question 4 Maths textbook Solution.

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Answer:-    x+y=\tan \left ( x+c \right )

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

Hint : - Differential equation of the form \frac{dy}{dx}=\int \left ( ax+by+c \right )can be reduced to variable separable form by substitution ax+by+c=v

Solution : -  We have,

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

Let x+y=v

Differentiating with respect to x, we get,

                            \begin{aligned} \; \; \; \; \; \; &\quad \frac{d }{d x}(\mathrm{x}+\mathrm{y})=\frac{d}{d x}(v) \\ &\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 the value of \frac{dy}{dx} in equation (i), we get

\frac{d v}{d x}-1=(\mathrm{x}+\mathrm{y})^{2} \\

\Rightarrow \frac{d v}{d x}-1=\mathrm{v}^{2}                                            (\because \mathrm{x}+\mathrm{y}=\mathrm{v}) \\

\Rightarrow \frac{d v}{d x}=\mathrm{v}^{2}+1 \\

\Rightarrow \frac{d v}{v^{2}+1}=\mathrm{d} \mathrm{x}

Integrating on both the sides, we get,

\int \frac{d v}{v^{2}+1}=\int \mathrm{d} \mathrm{x}

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

Putting v=x+y, we get

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

            \Rightarrow x+y=\tan \left ( x+c \right )

                        (This is the required solution).


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