Question

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

Answer 1

as we know that slope of tangent is

$\frac{dy}{dx}$

it is given that

$\frac{dy}{dx}=\frac{2x}{y^2}$

$y^{2}dy=2xdx$

Integrating both sides
We get,

$\int y^{2}dy=\int 2xdx$

$\frac{y^3}{3}=\frac{2x^2}{2}+c$

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

Put $C_1=3c$

$y^3=3x^2+C_1$ (i)

Given that equation passes through $(-2,3)$

Putting $x=-2,y=3$ in $\left(i\right)$ then we get,

$3(-2{)}^2+C_1=3^3$

$3\times 4+C_1=27$

$C_1=27-12$

$C_1=15$

Putting $C_1$ in $\left(i\right)$ then we get,

$y^3=3x^2+15$

$y=(3x^2+15{)}^{\frac{1}{3}}$

Final Answer:
Hence, the required equation of the curve is

$y=(3x^2+15{)}^{\frac{1}{3}}$