Consider the differential equation

\[\frac{{{\text{d}}y}}{{{\text{d}}x}} = f\left( {\frac{y}{x}} \right),{\text{ }}x > 0.\]

Use the substitution \(y = vx\) to show that the general solution of this differential equation is

\[\int {\frac{{{\text{d}}v}}{{f(v) - v}} = \ln x + } {\text{ Constant.}}\]

Hence, or otherwise, solve the differential equation

\[\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{{x^2} + 3xy + {y^2}}}{{{x^2}}},{\text{ }}x > 0,\]

given that \(y = 1\) when \(x = 1\). Give your answer in the form \(y = g(x)\).

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

the differential equation becomes

\(v + x\frac{{{\text{d}}v}}{{{\text{d}}x}} = f(v)\)     A1

\(\int {\frac{{{\text{d}}v}}{{f(v) - v}} = \int {\frac{{{\text{d}}v}}{x}} } \)     A1

integrating, Constant \(\int {\frac{{{\text{d}}v}}{{f(v) - v}} = \ln x + } {\text{ Constant}}\)     AG

[3 marks]

EITHER

\(f(v) = 1 + 3v + {v^2}\)     (A1)

\(\left( {\int {\frac{{{\text{d}}v}}{{f(v) - v}} = } } \right)\,\,\,\int {\frac{{{\text{d}}v}}{{1 + 3v + {v^2} - v}} = \ln x + C} \)     M1A1

\(\int {\frac{{{\text{d}}v}}{{{{(1 + v)}^2}}} = (\ln x + C)} \)     A1

 

Note:     A1 is for correct factorization.

 

\( - \frac{1}{{1 + v}}\,\,\,( = \ln x + C)\)     A1

OR

\(v + x\frac{{{\text{d}}v}}{{{\text{d}}x}} = 1 + 3v + {v^2}\)     A1

\(\int {\frac{{{\text{d}}v}}{{1 + 2v + {v^2}}} = \int {\frac{1}{x}{\text{d}}x} } \)     M1

\(\int {\frac{{{\text{d}}v}}{{{{(1 + v)}^2}}}\,\,\,\left( { = \int {\frac{1}{x}{\text{d}}x} } \right)} \)     (A1)

 

Note:     A1 is for correct factorization.

 

\( - \frac{1}{{1 + v}} = \ln x( + C)\)     A1A1

THEN

substitute \(y = 1\) or \(v = 1\) when \(x = 1\)     (M1)

therefore \(C = - \frac{1}{2}\)     A1

 

Note:     This A1 can be awarded anywhere in their solution.

 

substituting for \(v\),

\( - \frac{1}{{\left( {1 + \frac{y}{x}} \right)}} = \ln x - \frac{1}{2}\)     M1

 

Note:     Award for correct substitution of \(\frac{y}{x}\) into their expression.

 

\(1 + \frac{y}{x} = \frac{1}{{\frac{1}{2} - \ln x}}\)     (A1)

 

Note:     Award for any rearrangement of a correct expression that has \(y\) in the numerator.

 

\(y = x\left( {\frac{1}{{\left( {\frac{1}{2} - \ln x} \right)}} - 1} \right)\,\,\,{\text{(or equivalent)}}\)     A1

\(\left( { = x\left( {\frac{{1 + 2\ln x}}{{1 - 2\ln x}}} \right)} \right)\)

[10 marks]

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