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Need solution for RD Sharma maths class 12 chapter Differential Equations exercise 21.7 question 38 sub question (i)

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Answer: (y-x)-(\log (x(1+y)))=c

Hint: Separate the terms of x and y and then integrate them.

Given: x y \frac{d y}{d x}=1+x+y+x y

Solution: x y \frac{d y}{d x}=1+x+y+x y

        \begin{aligned} &x y \frac{d y}{d x}=(1+x)+y(1+x) \\\\ &x y \frac{d y}{d x}=(1+x)(1+y) \\\\ &\frac{y}{1+y} d y=\frac{1+x}{x} d x \end{aligned}

          Integrating both sides

        \begin{aligned} &\int 1 d y-\int \frac{1}{1+y} d y=\int \frac{1}{x} d x+\int 1 d x \\\\ &y-\log |1+y|=\log |x|+x+c \end{aligned}

        \begin{aligned} &y-x-\log |1+y|-\log |x|=c \\\\ &(y-x)-\log [x(1+y)]=c \end{aligned}

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