The variables z and x are related by the differential equation $$3z^2 \frac{d^2z}{dx^2} + 6z \frac{dz}{dx} + 6z^2 + 5z^2 \frac{dx}{dx} = 5x + 2.$$ Use the substitution $y = x^2$ to show that y and x are related by the differential equation $$\frac{dy}{dx^2} + 2 \frac{dy}{dx} = 5y + 2.$$ Given that $z = 1$ and \(\frac{dz}{dx} = -\frac{2}{3}\) when $x = 0$, find z in terms of x - CIE - A-Level Further Maths - Question 11 - 2010 - Paper 1

To find z in terms of x, we integrate using the conditions provided:

1. Start with the differential equation and substitute z = 1 and (\frac{dz}{dx} = -\frac{2}{3}).
2. Solve the resulting equation for z, taking into account the integration constants that arise from the conditions. The result is:

$z = \left[ e^{\left(4cos(2x) + 3sin(2x)\right)} \right] \cdot x^{1/3}$

As $x$ grows large, the exponentials in the earlier found equation become negligible. Hence, we can deduce:

$z \approx x^{1/3}$
as terms with lower powers of x contribute less significantly.

Calculate the limits as x approaches both 1 and infinity:

1. Vertical asymptote at x = 1 (denominator goes to 0).
2. As x goes to infinity, the function approaches y = 1 as it stabilizes.

Set the equation equal to the asymptote equation, solve:

$y = \frac{x(x + 1)}{(x - 1)^2} = 1$
This gives a single valid intersection point, confirmed via substitution.

First, differentiate the curve C:

$\frac{dy}{dx} = \frac{(x + 1)(x - 1)^2 - x(x + 1)2(x - 1)}{(x - 1)^4}$.
Set the numerator to zero to determine stationary points and subsequently find their coordinates.

Identify the intervals where the derived (\frac{dy}{dx} < 0). This requires analyzing the sign of the derivative and results in the relevant x intervals.

Sketch the curve showing the left-hand branch passing through the origin, the position of the asymptote, and validate the coordinate intersections found.