Find the equation of the curve passing through the point (-2, 3), given that the slope of the tangent to the curve at any point $(x, y)$ is $\dfrac{2x}{y^2}$

The correct answer is $y^3 = 3x^2 + 15$

Given that the slope is $\dfrac{2x}{y^2}$

⇒ $\dfrac{dy}{dx}$ = $\dfrac{2x}{y^2}$

⇒ $y^2 dy$ = $2x dx$


Integrating both sides,

⇒ $\int y^2 dy$ = $\int 2x dx$


$\dfrac{y^3}{3}$ = $\dfrac{2x^2}{2}$ + $C_1$

$y^3$ = $3x^2$ + $C$



For the point (-2, 3),

$(-3)^3$ = $3(-2)^2$ + $C$ ⇒ $C$ = $15$


The equation of the curve is $y^3 = 3x^2 + 15$