Sketch, on one diagram, the four isoclines corresponding to \(f(x,{\text{ }}y) = k\) where \(k\) takes the values \(-1\), \(-0.5\), \(0\) and \(1\). Indicate clearly where each isocline crosses the \(y\) axis.

A curve, \(C\), passes through the point \((0,1)\) and satisfies the differential equation above.

Sketch \(C\) on your diagram.

A curve, \(C\), passes through the point \((0,1)\) and satisfies the differential equation above.

State a particular relationship between the isocline \(f(x,{\text{ }}y) =  - 0.5\) and the curve \(C\), at their point of intersection.

A curve, \(C\), passes through the point \((0,1)\) and satisfies the differential equation above.

Use Euler’s method with a step interval of \(0.1\) to find an approximate value for \(y\) on \(C\), when \(x = 0{\text{.}}5\).

A1 for 4 parallel straight lines with a positive gradient     A1

A1 for correct \(y\) intercepts     A1

[2 marks]

A1 for passing through \((0,1)\) with positive gradient less than \(2\)

A1 for stationary point on \(y = 2x\)

A1 for negative gradient on both of the other \(2\) isoclines     A1A1A1

[3 marks]

The isocline is perpendicular to \(C\)     R1

[1 mark]

\({y_{n + 1}} = {y_n} + 0.1({y_n} - 2{x_n})\;\;\;( = 1.1{y_n} - 0.2{x_n})\)     (M1)(A1)

Note:     Also award M1A1 if no formula seen but \({y_2}\) is correct.

\({y_0} = 1,{\text{ }}{y_1} = 1.1,{\text{ }}{y_2} = 1.19,{\text{ }}{y_3} = 1.269,{\text{ }}{y_4} = 1.3359\)     (M1)

\({y_5} = 1.39{\text{ to 3sf}}\)     A1

Note:     M1 is for repeated use of their formula, with steps of \(0.1\).

Note:     Accept \(1.39\) or \(1.4\) only.

[4 marks]

Total [10 marks]

Some candidates ignored the instruction to prove from first principles and instead used standard differentiation. Some candidates also only found a derivative from one side.

Parts (b) and (c) were attempted by very few candidates. Few recognized that the gradient of the curve had to equal the value of \(k\) on the isocline.

Parts (b) and (c) were attempted by very few candidates. Few recognized that the gradient of the curve had to equal the value of \(k\) on the isocline.

Those candidates who knew the method managed to score well on this part. On most calculators a short program can be written in the exam to make Euler’s method very quick. Quite a few candidates were losing time by calculating and writing out many intermediate values, rather than just the \(x\) and\(y\) values.