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 \(y = x + c\)