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

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