Date | November 2019 | Marks available | 10 | Reference code | 19N.3.AHL.TZ0.Hca_4 |
Level | Additional Higher Level | Paper | Paper 3 | Time zone | Time zone 0 |
Command term | Solve | Question number | Hca_4 | Adapted from | N/A |
Question
Consider the differential equation dydx=4x2+y2−xyx2, 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.
Sketch the isoclines for dydx=4.
Express m2−2m+4 in the form (m−a)2+b , where a, b∈Z.
Solve the differential equation, for x>0, giving your answer in the form y=f(x).
Sketch the graph of y=f(x) for 1⩽x⩽1.4 .
With reference to the curvature of your sketch in part (c)(iii), and without further calculation, explain whether you conjecture f(1.4) will be less than, equal to, or greater than your answer in part (a).
Markscheme
(M1)(A1)(A1)(A1)A1
y(1.4)≈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 4x2+y2−xyx2=4 (M1)
⇒y2−xy=0
y(y−x)=0
y=0 or y=x
A1A1
[3 marks]
m2−2m+4=(m−1)2+3(a=1,b=3) A1
[1 mark]
recognition of homogeneous equation,
let y=vx M1
the equation can be written as
v+xdvdx=4+v2−v (A1)
xdvdx=v2−2v+4
∫1v2−2v+4dv=∫1xdx M1
Note: Award M1 for attempt to separate the variables.
∫1(v−1)2+3dv=∫1xdx from part (c)(i) M1
1√3arctan(v−1√3)=lnx(+c) A1A1
x=1,y=2⇒v=2
1√3arctan(1√3)=ln1+c M1
Note: Award M1 for using initial conditions to find c.
⇒c=π6√3(=0.302) A1
arctan(v−1√3)=√3lnx+π6
substituting v=yx M1
Note: This M1 may be awarded earlier.
y=x(√3tan(√3lnx+π6)+1) A1
[10 marks]
curve drawn over correct domain A1
[1 mark]
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(1.4) > estimate in part (a) R1
Note: The R1 is dependent on the A1.
[2 marks]