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Explain Solution R.D.Sharma Class 12 Chapter 21 Differential Equations Exercise 21.8 Question 9 Maths Textbook Solution.

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Answer: \frac{1}{2}\left ( y-x \right )+\frac{1}{2}log|x+y|=c

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

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

Solution : -  We have,

                     (x+y)(d x-d y)=d x+d y\\

             \Rightarrow x d x-x d y+y d x-y d y=d x+d y\\

            \Rightarrow x d x+y d x-d x=d y+x d y+y d y\\

            \Rightarrow(x+y-1) d x=(1+x+y) d y\\

            \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{x}+\mathrm{y}-1}{\mathrm{x}+\mathrm{y}+1}

Let x+y=v

Differentiating with respect to x, we get,

                            \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 \quad \frac{d y}{d x}=\frac{d v}{d x}-1

Substituting (ii) in equation (i), we get,

                \therefore \frac{d v}{d x}-1=\frac{x+y-1}{x+y+1} \\

            \Rightarrow \frac{d v}{d x}-1=\frac{v-1}{v+1} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \quad[\because x+y=v] \\

            \Rightarrow \frac{d v}{d x}=\frac{v-1}{v+1}+1 \\

             \Rightarrow \frac{d v}{d x}=\frac{v-1+v+1}{v+1} \\

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

                    Taking like variables on the same side, we get,

            \Rightarrow \frac{v+1}{2v}dv=dx

Integrating on both sides, we get,

                    \begin{aligned} &\Rightarrow \int \frac{v+1}{2 v} d v=\int d \mathrm{x} \\ &\Rightarrow \frac{1}{2} \int\left(1+\frac{1}{v}\right) d v=\int \mathrm{d} \mathrm{x} \\ &\Rightarrow \frac{1}{2} \int(\mathrm{v}+\log |v|)=\mathrm{x}+\mathrm{c} \end{aligned}

Putting v = x + y, we get,

                  \begin{aligned} &\Rightarrow \frac{1}{2}(x+y+\log |x+y|)=x+c \\ &\Rightarrow \quad(x+y+\log |x+y|)=2(x+c) \\ &\Rightarrow(x+y+\log |x+y|)=2 x+2 c \\ &\Rightarrow y-x+\log |x+y|=2 c \\ &\Rightarrow \frac{1}{2}(y-x)+\frac{1}{2} \log |x+y|=c \end{aligned}

                                (This is the required solution).

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