Question

The gradient of the tangent to a curve is given by dy/dx = y / x(x+1). The point P(3,6) lies on the curve.

a) Find the equation of the tangent to the curve at P.

b) Solve the differential equation to find the equation of the curve in the form y=f(x).

Answer 1

a) Slope at the point (3,6): $y'(3)=\frac{6}{3*4}=\frac{1}{2}$

Equation of the tangent: $(y-6)=\frac{1}{2}(x-3)$ or $y=\frac{1}{2}x-\frac{9}{2}$

b) $\frac{dy}{dx}=\frac{y}{x(x+1)}$

$\frac{dy}{y}=\frac{dx}{x(x+1)}$

$\frac{dy}{y}=(\frac{1}{x}-\frac{1}{x+1})dx$

$\int\frac{dy}{y}=\int(\frac{1}{x}-\frac{1}{x+1})dx$

$lny=lnx-ln(x+1)+C$

$lny=ln\frac{x}{x+1}+C$

$y=\frac{Dx}{x+1}$

$y(3)=6\to6=\frac{3D}{4}\to D=8$

Thus, $y=\frac{8x}{x+1}$