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Provide solution for  RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Revision Exercise (RE) Question 63 textbook solution.

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Answer :    y=\frac{1}{2 x^{2}+1}

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

Given            :   \frac{d y}{d x}=-4 x y^{2}\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad y=1, x=0

Solution        : \frac{d y}{d x}=-4 x y^{2}

                       \frac{1}{y^{2}}dy=-4x \; dx

Integrating  both sides ,

              \begin{gathered} \int \frac{1}{y^{2}} d y=-4 \int x d x \\ \Rightarrow \frac{-1}{y}=-2 x^{2}+c \end{gathered}

Now, x=0, y=1

Therefore, -1= 0+c

                      c= -1

Put value of c,

\begin{aligned} &\frac{-1}{y}=-2 x^{2}-1 \\ &\frac{1}{y}=2 x^{2}+1 \\ &y=\frac{1}{2 x^{2}+1} \end{aligned}

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