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  • Explain solution RD Sharma class 12 chapter Differential Equations exercise 21.7 question 4 maths

Explain solution RD Sharma class 12 chapter Differential Equations exercise 21.7 question 4 maths

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Answer: \log |y|=\frac{2}{3} x^{3}+x^{2}+2 x+2 \log |x-1|+c

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

Given: (x-1) \frac{d y}{d x}=2 x^{3} y

Solution: (x-1) \frac{d y}{d x}=2 x^{3} y

        \begin{gathered} \frac{d y}{y}=\frac{2 x^{8}}{x-1} d x \\\\ \frac{d y}{y}=\frac{2\left((x-1)\left(x^{2}+x+1\right)+1\right)}{(x-1)} d x \\\\ \frac{d y}{y}=2\left(x^{2}+x+1+\frac{1}{(x-1)}\right) d x \end{gathered}

Integrating both sides

        \begin{aligned} &\int \frac{d y}{y}=\int 2\left(x^{2}+x+1+\frac{1}{(x-1)}\right) d x \\\\ &\log |y|=\frac{2 x^{3}}{3}+\frac{2 x^{2}}{2}+2 x+2 \log |x-1|+c \\\\ &\log |y|=\frac{2 x^{3}}{3}+x^{2}+2 x+2 \log |x-1|+c \end{aligned}

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