A curve for which $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{x^3+a}{x^2}$, where $a$ is a con

Question

A curve for which $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{x^3+a}{x^2}$, where $a$ is a constant, has a gradient value of $4$ at the point $(-1, 3)$. Find the value of $a$. Find the equation of the curve.

Accepted Answer

(i) $a=5$ (ii) $y=\frac{1}{2}x^2-\frac{5}{x}-\frac{5}{2}$

Integration - Finding the Equation of the Curve

Integration - Finding the Equation of the Curve

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