17M.3.AHL.TZ0.HCA_4

pestleMathematicsAAHLPaper 317M· ahl-5-18-1st-order-des-euler-method-variables-separable-integrating-factor-homogeneous-de-using-sub-y=vxsource ↗

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

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

[3]

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 $y = g(x)$ .

[10]

Markscheme / solution

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

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

$v + x\frac{{{\text{d}}v}}{{{\text{d}}x}} = 1 + 3v + {v^2}$ 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]

Examiners’ report

[N/A]