Consider the differential equation $\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{y}{{x + \sqrt {xy} }}$, for $x,{\text{ }}y > 0$.

(a)     Use Euler’s method starting at the point $(x,{\text{ }}y) = (1,{\text{ }}2)$, with interval $h = 0.2$, to find an approximate value of y when $x = 1.6$.

(b)     Use the substitution $y = vx$ to show that $x\frac{{{\text{d}}v}}{{{\text{d}}x}} = \frac{v}{{1 + \sqrt v }} - v$.

(c)     (i)     Hence find the solution of the differential equation in the form $f(x,{\text{ }}y) = 0$, given that $y = 2$ when $x = 1$.

(ii)     Find the value of $y$ when $x = 1.6$.

(a)     let $f(x,{\text{ }}y) = \frac{y}{{x + \sqrt {xy} }}$

$y(1.2) = y(1) + 0.2f(1,{\text{ }}2){\text{ }}( = 2 + 0.1656 \ldots )$     (M2)(A1)

$= 2.1656 \ldots$     A1

$y(1.4) = 2.1656 \ldots  + 0.2f(1.2,{\text{ }}2.1256 \ldots ){\text{ }}( = 2.1656 \ldots  + 0.1540 \ldots )$     (M1)

 

Note: M1 is for attempt to apply formula using point $\left( {1.2,{\text{ }}y(1.2)} \right)$.

 

$= 2.3197 \ldots$     A1

$y(1.6) = 2.3197 \ldots  + 0.2f(1.4,{\text{ }}2.3197 \ldots ){\text{ }}( = 2.3297 \ldots  + 0.1448 \ldots )$

$= 2.46$   (3sf)     A1     N3

[7 marks]

(b)     $y = vx \Rightarrow \frac{{{\text{d}}y}}{{{\text{d}}x}} = v + x\frac{{{\text{d}}v}}{{{\text{d}}x}}$     (M1)

$\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{y}{{x + \sqrt {xy} }} \Rightarrow v + x\frac{{{\text{d}}v}}{{{\text{d}}x}} = \frac{{vx}}{{x + \sqrt {v{x^2}} }}$     M1

$\Rightarrow v + x\frac{{{\text{d}}v}}{{{\text{d}}x}} = \frac{{vx}}{{x + x\sqrt v }}{\text{ (as }}x > 0)$     A1

$\Rightarrow x\frac{{{\text{d}}v}}{{{\text{d}}x}} = \frac{v}{{1 + \sqrt v }} - v$     AG

[3 marks]

(c)     (i)     $x\frac{{{\text{d}}v}}{{{\text{d}}x}} = \frac{v}{{1 + \sqrt v }} - v$

$x\frac{{{\text{d}}v}}{{{\text{d}}x}} = \frac{{ - v\sqrt v }}{{1 + \sqrt v }} \Rightarrow \frac{{1 + \sqrt v }}{{ - v\sqrt v }}{\text{d}}v = \frac{1}{x}{\text{d}}x$     M1

$\int {\frac{{1 + \sqrt v }}{{ - v\sqrt v }}{\text{d}}v = \int {\frac{1}{x}{\text{d}}x} }$     (M1)

$\frac{2}{{\sqrt v }} - \ln v = \ln x + C$     A1A1

 

Note: Do not penalize absence of $+ C$ at this stage; ignore use of absolute values on \(v\)and \(x\) (which are positive anyway).

 

$2\sqrt {\frac{x}{y}}  - \ln \frac{y}{x} = \ln x + C$ as $y = vx \Rightarrow v = \frac{y}{x}$     M1

$y = 2$ when $x = 1 \Rightarrow \sqrt 2  - \ln 2 = 0 + C$     M1

$2\sqrt {\frac{x}{y}}  - \ln \frac{y}{x} = \ln x + \sqrt 2  - \ln 2$

$2\sqrt {\frac{x}{y}}  - \ln \frac{y}{x} - \ln x - \sqrt 2  + \ln 2 = 0$   $\left( {2\sqrt {\frac{x}{y}}  - \ln y - \sqrt 2  + \ln 2 = 0} \right)$     A1

(ii)     $2\sqrt {\frac{{1.6}}{y}}  - \ln \frac{y}{{1.6}} - \ln 1.6 - \sqrt 2  + \ln 2 = 0$     (M1)

$y = 2.45$     A1

[9 marks]

Part (a) was well answered by most candidates. In a few cases calculation errors and early rounding errors prevented candidates from achieving full marks, but most candidates scored at least a few marks here. In part (b) some candidates failed to convincingly show the given result. Part (c) proved to be a hard question for many candidates and a significant number of candidates had difficulty manipulating the algebraic expression, and either had the incorrect expression to integrate, or incorrectly integrated the correct expression. Many candidates reached as far as separating the variables correctly but integrating proved to be too difficult for many of them although most realised that the expression on v had to be split into two separate integrals. Most candidates made good attempts to evaluate the arbitrary constant and arrived at a correct or almost correct expression (sign errors were a common error) which allowed follow through for part b (ii). In some cases however the expression obtained was too simple or was omitted and it was not possible to grant follow through marks.