The variables x and y satisfy the differential equation \(\frac{dy}{dx} = \frac{y^2 + 4}{x(y + 4)}\) for \(x > 0\). It is given that \(x = 4\) when \(y = 2\sqrt{3}\). Solve the differential equation to obtain the value of \(x\) when \(y = 2\).

The variables x and y are related by the differential equation \((x^2 + 4) \frac{dy}{dx} = 6xy\).

It is given that \(y = 32\) when \(x = 0\). Find an expression for y in terms of x.

Given that \(x = 1\) when \(t = 0\), solve the differential equation

\(\frac{dx}{dt} = \frac{1}{x} - \frac{x}{4}\)

obtaining an expression for \(x^2\) in terms of \(t\).

Given that \(y = 0\) when \(x = 1\), solve the differential equation \(xy \frac{dy}{dx} = y^2 + 4\), obtaining an expression for \(y^2\) in terms of \(x\).

Given that \(y = 2\) when \(x = 0\), solve the differential equation

\(\frac{y}{\frac{dy}{dx}} = 1 + y^2,\)

obtaining an expression for \(y^2\) in terms of \(x\).

Given that \(y = 1\) when \(x = 0\), solve the differential equation \(\frac{dy}{dx} = \frac{y^3 + 1}{y^2}\), obtaining an expression for \(y\) in terms of \(x\).

The variables x and y satisfy the differential equation \(\frac{dy}{dx} = \frac{xy}{1+x^2}\), and \(y = 2\) when \(x = 0\). Solve the differential equation, obtaining a simplified expression for y in terms of x.

A curve is such that the gradient at a general point with coordinates \((x, y)\) is proportional to \(\frac{y}{\sqrt{x+1}}\).

The curve passes through the points with coordinates \((0, 1)\) and \((3, e)\).

By setting up and solving a differential equation, find the equation of the curve, expressing \(y\) in terms of \(x\).

The coordinates (x, y) of a general point of a curve satisfy the differential equation \(x \frac{dy}{dx} = (1 - 2x^2)y\), for \(x > 0\). It is given that \(y = 1\) when \(x = 1\).

Solve the differential equation, obtaining an expression for \(y\) in terms of \(x\).

The variables x and y satisfy the differential equation \((x + 1) y \frac{dy}{dx} = y^2 + 5\).

It is given that \(y = 2\) when \(x = 0\). Solve the differential equation obtaining an expression for \(y^2\) in terms of \(x\).

The coordinates \((x, y)\) of a general point on a curve satisfy the differential equation \(x \frac{dy}{dx} = (2 - x^2)y\).

The curve passes through the point \((1, 1)\). Find the equation of the curve, obtaining an expression for \(y\) in terms of \(x\).

The variables x and y satisfy the differential equation

\((x+1) \frac{dy}{dx} = y(x+2),\)

and it is given that \(y = 2\) when \(x = 1\). Solve the differential equation and obtain an expression for \(y\) in terms of \(x\).

The variables x and y satisfy the differential equation

\(x \frac{dy}{dx} = y(1 - 2x^2)\),

and it is given that \(y = 2\) when \(x = 1\). Solve the differential equation and obtain an expression for y in terms of x in a form not involving logarithms.

The variables x and t satisfy the differential equation

\(t \frac{dx}{dt} = \frac{k - x^3}{2x^2}\),

for \(t > 0\), where \(k\) is a constant. When \(t = 1, x = 1\) and when \(t = 4, x = 2\).

(i) Solve the differential equation, finding the value of \(k\) and obtaining an expression for \(x\) in terms of \(t\). [9]

(ii) State what happens to the value of \(x\) as \(t\) becomes large. [1]