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  • Explain solution RD Sharma class 12 chapter Differential Equations exercise 21.7 question 12 maths

Explain solution RD Sharma class 12 chapter Differential Equations exercise 21.7 question 12 maths

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Answer: y-x=\log |x|-\log |y-1|+c

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

Given: x y d y=(y-1)(x+1) d x

Solution: x y d y=(y-1)(x+1) d x

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

        Integrating both sides

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

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