Consider the differential equation \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{{y^2} + {x^2}}}{{2{x^2}}}\) for which y = −1 when x = 1.

(a)     Use Euler’s method with a step length of 0.25 to find an estimate for the value of y when x = 2 .

(b)     (i)     Solve the differential equation giving your answer in the form \(y = f(x)\) .

         (ii)     Find the value of y when x = 2 .

(a)     Using an increment of 0.25 in the x-values     A1

Note: The A1 marks are awarded for final column.

\( \Rightarrow y(2) \approx - 0.304\)     A1

[7 marks]

(b)     (i)     let y = vx     M1

\( \Rightarrow \frac{{{\text{d}}y}}{{{\text{d}}x}} = v + x\frac{{{\text{d}}v}}{{{\text{d}}x}}\)     (A1)

\( \Rightarrow v + x\frac{{{\text{d}}v}}{{{\text{d}}x}} = \frac{{{v^2}{x^2} + {x^2}}}{{2{x^2}}}\)     (M1)

\( \Rightarrow x\frac{{{\text{d}}v}}{{{\text{d}}x}} = \frac{{1 - 2v + {v^2}}}{2}\)     (A1)

\( \Rightarrow x\frac{{{\text{d}}v}}{{{\text{d}}x}} = \frac{{{{(1 - v)}^2}}}{2}\)     A1

\( \Rightarrow \int {\frac{2}{{{{(1 - v)}^2}}}{\text{d}}v = \int {\frac{1}{x}{\text{d}}x} } \)     M1

\( \Rightarrow 2{(1 - v)^{ - 1}} = \ln x + c\)     A1A1

\( \Rightarrow \frac{2}{{1 - \frac{y}{x}}} = \ln x + c\)

when \(x = 1,{\text{ }}y = - 1 \Rightarrow c = 1\)     M1A1

\( \Rightarrow \frac{{2x}}{{x - y}} = \ln x + 1\)

\( \Rightarrow y = x - \frac{{2x}}{{1 + \ln x}}{\text{ }}\left( { = \frac{{x\ln x - x}}{{1 + \ln x}}} \right)\)     M1A1

(ii)     when \(x = 2,{\text{ }}y = - 0.362\,\,\,\,\,\left( {{\text{accept 2}} - \frac{4}{{1 + \ln 2}}} \right)\)     A1

[13 marks]

Total [20 marks]

Part (a) was well done by many candidates, but a number were penalised for not using a sufficient number of significant figures. Part (b) was started by the majority of candidates, but only the better candidates were able to reach the end. Many were unable to complete the question correctly because they did not know what to do with the substitution y = vx and because of arithmetic errors and algebraic errors.