In the diagram, the tangent to a curve at the point \(P\) with coordinates \((x, y)\) meets the \(x\)-axis at \(T\). The point \(N\) is the foot of the perpendicular from \(P\) to the \(x\)-axis. The curve is such that, for all values of \(x\), the gradient of the curve is positive and \(TN = 2\).

(i) Show that the differential equation satisfied by \(x\) and \(y\) is \(\frac{dy}{dx} = \frac{1}{2}y\).

The point with coordinates \((4, 3)\) lies on the curve.

(ii) Solve the differential equation to obtain the equation of the curve, expressing \(y\) in terms of \(x\).