19M.3.AHL.TZ0.HCA_5

pestleMathematicsAAHLPaper 319M· ahl-5-18-1st-order-des-euler-method-variables-separable-integrating-factor-homogeneous-de-using-sub-y=vxsource ↗

Consider the differential equation $2xy\frac{{{\text{d}}y}}{{{\text{d}}x}} = {y^2} - {x^2}$ , where $x > 0$ .

Solve the differential equation and show that a general solution is ${x^2} + {y^2} = cx$ where $c$ is a positive constant.

[11]

Prove that there are two horizontal tangents to the general solution curve and state their equations, in terms of $c$ .

[5]

Markscheme / solution

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

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

let $y = vx$ M1

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

$v + x\frac{{{\text{d}}v}}{{{\text{d}}x}} = \frac{{{v^2}{x^2} - {x^2}}}{{2v{x^2}}}$ (M1)

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

Note: Or equivalent attempt at simplification.

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

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

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

${\text{ln}}\left( {1 + {v^2}} \right) = - {\text{ln}}\,x + {\text{ln}}\,c$ A1A1

Note: Award A1 for LHS and A1 for RHS and a constant.

${\text{ln}}\left( {1 + {{\left( {\frac{y}{x}} \right)}^2}} \right) = - {\text{ln}}\,x + {\text{ln}}\,c$ M1

Note: Award M1 for substituting $v = \frac{y}{x}$ . May be seen at a later stage.

$1 + {\left( {\frac{y}{x}} \right)^2} = \frac{c}{x}$ A1

Note: Award A1 for any correct equivalent equation without logarithms.

${x^2} + {y^2} = cx$ AG

[11 marks]

METHOD 1

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

(for horizontal tangents) $\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0$ M1

$\left( { \Rightarrow {y^2} = {x^2}} \right) \Rightarrow y = \pm x$

EITHER

using ${x^2} + {y^2} = cx \Rightarrow 2{x^2} = cx$ M1

$2{x^2} - cx = 0 \Rightarrow x = \frac{c}{2}$ A1

Note: Award M1A1 for $2{y^2} = \pm cy$ .

using implicit differentiation of ${x^2} + {y^2} = cx$

$2x + 2y\frac{{{\text{d}}y}}{{{\text{d}}x}} = c$ M1

Note: Accept differentiation of $y = \sqrt {cx - {x^2}}$ .

$\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0 \Rightarrow x = \frac{c}{2}$ A1

THEN

tangents at $y = \frac{c}{2},\,\,y = - \frac{c}{2}$ A1A1

hence there are two tangents AG

METHOD 2

${x^2} + {y^2} = cx$

${\left( {x - \frac{c}{2}} \right)^2} + {y^2} = \frac{{{c^2}}}{4}$ M1A1

this is a circle radius ${\frac{c}{2}}$ centre $\left( {\frac{c}{2},\,\,0} \right)$ A1

hence there are two tangents AG

tangents at $y = \frac{c}{2},\,\,y = - \frac{c}{2}$ A1A1

[5 marks]

Examiners’ report

[N/A]