Question 19
**(b)** The variables $x$ and $y$ satisfy the differential equation
$\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{x^3+5x^2-2x-15}{6y(x^2-3)}.$
It is given that $y = 2$ when $x = 2$.
Solve the differential equation to obtain an expression for $y^2$ in terms of $x$.
(In part (a) you found $x^3+5x^2-2x-15 = (x+5)(x^2-3)+x$.)
$\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{x^3+5x^2-2x-15}{6y(x^2-3)}.$
It is given that $y = 2$ when $x = 2$.
Solve the differential equation to obtain an expression for $y^2$ in terms of $x$.
(In part (a) you found $x^3+5x^2-2x-15 = (x+5)(x^2-3)+x$.)
