Solve this differential equation by separating the variables, giving your answer in the form y = f (x) .

Solve the same differential equation by using the standard homogeneous substitution y = vx .

Solve the same differential equation by the use of an integrating factor.

If y = 20 when x = 2 , find y when x = 5 .

$\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{y}{x} \Rightarrow \int {\frac{1}{y}{\text{d}}y = \int {\frac{1}{x}{\text{d}}x} }$     M1

$\Rightarrow \ln y = \ln x + c$     A1

$\Rightarrow \ln y = \ln x + \ln k = \ln kx$

$\Rightarrow y = kx$     A1

[3 marks]

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

so $v + x\frac{{{\text{d}}v}}{{{\text{d}}x}} = v$     M1

$\Rightarrow x\frac{{{\text{d}}v}}{{{\text{d}}x}} = 0 \Rightarrow \frac{{{\text{d}}v}}{{{\text{d}}x}} = 0\,\,\,\,\,({\text{as }}x \ne 0)$     R1

$\Rightarrow v = k$

$\Rightarrow \frac{y}{x} = k\,\,\,\,\,( \Rightarrow y = kx)$     A1

[4 marks]

$\frac{{{\text{d}}y}}{{{\text{d}}x}} + \left( {\frac{{ - 1}}{x}} \right)y = 0$     (M1)

${\text{IF}} = {{\text{e}}^{\int {\frac{{ - 1}}{x}{\text{d}}x} }} = {{\text{e}}^{ - \ln x}} = \frac{1}{x}$     M1A1

${x^{ - 1}}\frac{{{\text{d}}y}}{{{\text{d}}x}} - {x^{ - 2}}y = 0$

$\Rightarrow \frac{{{\text{d}}[{x^{ - 1}}y]}}{{{\text{d}}x}} = 0$     (M1)

$\Rightarrow {x^{ - 1}}y = k\,\,\,\,\,( \Rightarrow y = kx)$     A1

[5 marks]

$20 = 2k \Rightarrow k = 10{\text{  so  }}y(5) = 10 \times 5 = 50$     A1

[1 mark]

This question allowed candidates to demonstrate a range of skills in solving differential equations. Generally this was well done with candidates making mistakes in algebra rather than the techniques themselves. For example a common error in part (a) was to go from $\ln y = \ln x + c$ to $y = x + c$

This question allowed candidates to demonstrate a range of skills in solving differential equations. Generally this was well done with candidates making mistakes in algebra rather than the techniques themselves. For example a common error in part (a) was to go from $\ln y = \ln x + c$ to $y = x + c$

This question allowed candidates to demonstrate a range of skills in solving differential equations. Generally this was well done with candidates making mistakes in algebra rather than the techniques themselves. For example a common error in part (a) was to go from $\ln y = \ln x + c$ to $y = x + c$

This question allowed candidates to demonstrate a range of skills in solving differential equations. Generally this was well done with candidates making mistakes in algebra rather than the techniques themselves. For example a common error in part (a) was to go from $\ln y = \ln x + c$ to $y = x + c$