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

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

Answers (1)

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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