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Please solve RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Revision Exercise (RE) Question 69 textbook solution.

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Answer : y=\left(3 x^{2}+15\right)^{\frac{1}{3}}

Hint              : use variable separable method

Given           : slope of the tangent to the curve at any   (x, y) \text { is } \frac{2 x}{y^{2}}

Solution :  Slope of tangent =\frac{dy}{dx}

                  \begin{aligned} \frac{d y}{d x} &=\frac{2 x}{y^{2}} \\ \Rightarrow \quad y^{2} d y &=2 x d x \end{aligned}

Integrating both sides

         \begin{gathered} \int y^{2} d y=\int 2 x d x \\ \end{gathered}

          \begin{gathered} \frac{y^{3}}{3}=2 \frac{x^{2}}{2}+c \\ \end{gathered}

        \begin{gathered} \frac{y^{3}}{3}=x^{2}+c \\ \end{gathered}

      \begin{gathered} y^{3}=3 x^{2}+3 c \end{gathered}

        \begin{aligned} &y^{3}=3 x^{2}+c_{1} &.....\text { (i) where } c_{1}=3 c \end{aligned}

Given that equation passes through (-2, 3)

Putting x=-2, y=3 in (i)

                         \begin{aligned} &y^{3}=3 x^{2}+c_{1} \\ &3^{3}=3(-2)^{2}+c_{1} \\ &27=3 \times 4+c_{1} \quad \Rightarrow c_{1}=27-12=15 \end{aligned}

Putting C_{1} in (i)

              \begin{aligned} &y^{3}=3 x^{2}+15 \\ &y=\left(3 x^{2}+15\right)^{\frac{1}{3}} \end{aligned}           is the perticular solution of the equation

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