19N.3.AHL.TZ0.HCA_4

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

Consider the differential equation $\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{4{x^2} + {y^2} - xy}}{{{x^2}}}$ , with $y = 2$ when $x = 1$ .

Use Euler’s method, with step length $h = 0.1$ , to find an approximate value of $y$ when $x = 1.4$ .

[5]

Sketch the isoclines for $\frac{{{\text{d}}y}}{{{\text{d}}x}} = 4$ .

[3]

Express ${m^2} - 2m + 4$ in the form ${\left( {m - a} \right)^2} + b$ , where $a{\text{, }}b \in \mathbb{Z}$ .

[1]

c.i.

Solve the differential equation, for $x > 0$ , giving your answer in the form $y = f\left( x \right)$ .

[10]

c.ii.

Sketch the graph of $y = f\left( x \right)$ for $1 \leqslant x \leqslant 1.4$ .

[1]

c.iii.

With reference to the curvature of your sketch in part (c)(iii), and without further calculation, explain whether you conjecture $f\left( {1.4} \right)$ will be less than, equal to, or greater than your answer in part (a).

[2]

c.iv.

Markscheme / solution

(M1)(A1)(A1)(A1)A1

$y\left( {1.4} \right) \approx 5.34$

Note: Award A1 for each correct $y$ value.
For the intermediate $y$ values, accept answers that are accurate to 2 significant figures.
The final $y$ value must be accurate to 3 significant figures or better.

[5 marks]

attempt to solve $\frac{{4{x^2} + {y^2} - xy}}{{{x^2}}} = 4$ (M1)

$\Rightarrow {y^2} - xy = 0$

$y\left( {y - x} \right) = 0$

$y = 0$ or $y = x$

A1A1

[3 marks]

${m^2} - 2m + 4 = {\left( {m - 1} \right)^2} + 3\,\,\,\,\,\left( {a = 1,\,b = 3} \right)$ A1

[1 mark]

c.i.

recognition of homogeneous equation,
let $y = vx$ M1

the equation can be written as

$v + x\frac{{{\text{d}}v}}{{{\text{d}}x}} = 4 + {v^2} - v$ (A1)

$x\frac{{{\text{d}}v}}{{{\text{d}}x}} = {v^2} - 2v + 4$

$\int {\frac{1}{{{v^2} - 2v + 4}}} {\text{d}}v = \int {\frac{1}{x}} {\text{d}}x$ M1

Note: Award M1 for attempt to separate the variables.

$\int {\frac{1}{{{{\left( {v - 1} \right)}^2} + 4}}} {\text{d}}v = \int {\frac{1}{x}} {\text{d}}x$ from part (c)(i) M1

$\frac{1}{{\sqrt 3 }}{\text{arctan}}\left( {\frac{{v - 1}}{{\sqrt 3 }}} \right) = {\text{ln}}\,x\,\,\left( { + c} \right)$ A1A1

$x = 1{\text{,}}\,y = 2 \Rightarrow v = 2$

$\frac{1}{{\sqrt 3 }}{\text{arctan}}\left( {\frac{1}{{\sqrt 3 }}} \right) = {\text{ln}}\,1 + c$ M1

Note: Award M1 for using initial conditions to find $c$ .

$\Rightarrow c = \frac{\pi }{{6\sqrt 3 }}\,\,\left( { = 0.302} \right)$ A1

${\text{arctan}}\left( {\frac{{v - 1}}{{\sqrt 3 }}} \right) = \sqrt 3 \,{\text{ln}}\,x + \frac{\pi }{6}$

substituting $v = \frac{y}{x}$ M1

Note: This M1 may be awarded earlier.

$y = x\left( {\sqrt 3 {\text{tan}}\left( {\sqrt 3 \,{\text{ln}}\,x + \frac{\pi }{6}} \right) + 1} \right)$ A1

[10 marks]

c.ii.

curve drawn over correct domain A1

[1 mark]

c.iii.

the sketch shows that $f$ is concave up A1

Note: Accept ${f'}$ is increasing.

this means the tangent drawn using Euler’s method will give an underestimate of the real value, so $f\left( {1.4} \right)$ > estimate in part (a) R1

Note: The R1 is dependent on the A1.

[2 marks]

c.iv.

Examiners’ report

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

c.iii.

[N/A]

c.iv.