(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.