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Provide solution for  RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Revision Exercise (RE) Question 62 textbook solution.

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Answer : x^{2}+y^{2}+2 x-4 y+c=0

Hint               : You must know the rules of solving differential equation and integration

Given            :   \frac{d y}{d x}=\frac{x+1}{2-y}\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad, y \neq 2

Solution        : \frac{d y}{d x}=\frac{x+1}{2-y}

           (2-y) d y=(x+1) d x

Integrating both sides,

         \begin{array}{r} \int(2-y) d y=\int(x+1) d x \\ 2 y-\frac{y^{2}}{2}=\frac{x^{2}}{2}+x+c_{1} \\ \frac{x^{2}}{2}+x+c_{1}-2 y+\frac{y^{2}}{2}=0 \end{array}

    \begin{array}{ll} x^{2}+2 x+y^{2}-4 y+2 c_{1} & =0 \\ x^{2}+y^{2}+2 x-4 y+c & =0\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[2 c_{1}=c\right] \end{array}

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