Question

Find the equation of the curve passing through $(-2,3)$ on which the gradient of the tangent at any point $(x,y)$ is \[ \frac{dy}{dx} = \frac{2x}{y^2}. \]

Hint:

Always separate variables when the derivative is expressed as a rational function of $x$ and $y$. Don't forget to apply the given point to determine the constant of integration.

Answer 1

Step 1: Write the differential equation.
\[ \frac{dy}{dx} = \frac{2x}{y^2}. \]

Step 2: Separate variables.
\[ y^2 \, dy = 2x \, dx. \]

Step 3: Integrate both sides.
\[ \int y^2 \, dy = \int 2x \, dx. \] \[ \frac{y^3}{3} = x^2 + C. \] So, \[ y^3 = 3x^2 + C'. \]

Step 4: Use initial condition.
Point $(-2,3)$ lies on the curve. Substitute: \[ 3^3 = 3(-2)^2 + C' $\Rightarrow$ 27 = 12 + C' $\Rightarrow$ C' = 15. \]

Step 5: Final equation of curve.
\[ y^3 = 3x^2 + 15. \]

Final Answer: \[ \boxed{y^3 = 3x^2 + 15} \]