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

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

Answers (1)

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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