(a)     Show that the solution of the homogeneous differential equation

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{y}{x} + 1,{\text{ }}x > 0,\)

given that \(y = 0{\text{ when }}x = {\text{e, is }}y = x(\ln x - 1)\).

(b)     (i)     Determine the first three derivatives of the function \(f(x) = x(\ln x - 1)\).

  (ii)     Hence find the first three non-zero terms of the Taylor series for f(x) about x = 1.

(a)     EITHER

use the substitution y = vx

\(\frac{{{\text{d}}y}}{{{\text{d}}x}}x + v = v + 1\)     M1A1

\(\int {{\text{d}}v = \int {\frac{{{\text{d}}x}}{x}} } \)

by integration

\(v = \frac{y}{x} = \ln x + c\)     A1

OR

the equation can be rearranged as first order linear

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} - \frac{1}{x}y = 1\)     M1

the integrating factor I is

\({{\text{e}}^{\int { - \frac{1}{x}{\text{d}}x} }} = {{\text{e}}^{ - \ln x}} = \frac{1}{x}\)     A1

multiplying by I gives

\(\frac{{{\text{d}}}}{{{\text{d}}x}}\left( {\frac{1}{x}y} \right) = \frac{1}{x}\)

\(\frac{1}{x}y = \ln x + c\)     A1

THEN

the condition gives c = –1

so the solution is \(y = x(\ln x - 1)\)     AG

[5 marks]

(b)     (i)     \(f'(x) = \ln x - 1 + 1 = \ln x\)     A1

\(f''(x) = \frac{1}{x}\)     A1

\(f'''(x) = - \frac{1}{{{x^2}}}\)     A1

(ii)     the Taylor series about x = 1 starts

\(f(x) \approx f(1) + f'(1)(x - 1) + f''(1)\frac{{{{(x - 1)}^2}}}{{2!}} + f'''(1)\frac{{{{(x - 1)}^3}}}{{3!}}\)     (M1)

\( =  - 1 + \frac{{{{(x - 1)}^2}}}{{2!}} - \frac{{{{(x - 1)}^3}}}{{3!}}\)     A1A1A1

[7 marks]

Total: [12 marks]

Part(a) was well done by many candidates. In part(b)(i), however, it was disappointing to see so many candidates unable to differentiate \(x(\ln x - 1)\) correctly. Again, too many candidates were able to quote the general form of a Taylor series expansion, but not how to apply it to the given function.