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Explain Solution R.D.Sharma Class 12 Chapter 21 Differential Equations Exercise Revision Exercise Question 27 Maths Textbook Solution.

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Answer: x+y+log|xy|=c

Hint: Separate x & y and then integrate

Given:  \left ( 1+x \right )y\: dx+\left ( 1+y \right )x\: dy=0

Solution:  \left ( 1+x \right )y\: dx+\left ( 1+y \right )x\: dy=0

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

            Integrating both sides, we get

            \begin{aligned} &\int\left(\frac{1}{y}+1\right) d y=-\int\left(\frac{1}{x}+1\right) d x \\ &\Rightarrow \int \frac{1}{y} d y+\int d y=-\int \frac{1}{x} d x-\int d x \\ &\Rightarrow \log |y|+y=-\log |x|-x+c \\ &\Rightarrow \log |y|+\log |x|+x+y=c \\ &\Rightarrow x+y+\log |x y|=c(\because \log a+\log b=\log a b) \end{aligned}

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