\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 2\).

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = x + 1\).

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = x - 1\).

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \) constant.

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = f\left( x \right)\).

The slope field for the differential equation \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = x + y\) for \( - 4 \leqslant x \leqslant 4,\,\, - 4 \leqslant y \leqslant 4\) is shown in the following diagram.

Explain why the slope field indicates that the only linear solution is \(y =  - x - 1\).

Given that all the isoclines from a slope field of a differential equation are straight lines through the origin, find two examples of the differential equation.

     A2

[2 marks]

   A2

[2 marks]

  A2

[2 marks]

the slope is the same everywhere     A1

[1 mark]

all points that have the same \(x\) coordinate have the same slope    A1

[1 mark]

this is where a straight line appears on the slope field        A1

There is no other straight line, all the other solutions are curves        A1

[2 marks]

given \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = f\left( {x,\,y} \right)\), the isoclines are \(f\left( {x,\,y} \right) = k\)      (M1)

here the isoclines are \(y = kx\) (or \(x = ky\))     (A1)

any two differential equations of the correct form, for example

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{ky}}{x},\,\,\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{kx}}{y},\,\,\frac{{{\text{d}}y}}{{{\text{d}}x}} = {\text{sin}}\left( {\frac{y}{x}} \right),\,\,\frac{{{\text{d}}y}}{{{\text{d}}x}} = {\text{sin}}\left( {\frac{x}{y}} \right)\)      A1A1

[4 marks]