Date | November 2018 | Marks available | 6 | Reference code | 18N.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=1+yx, where x≠0.
Consider the family of curves which satisfy the differential equation dydx=1+yx, where x≠0.
Given that y(1)=1, use Euler’s method with step length h = 0.25 to find an approximation for y(2). Give your answer to two significant figures.
Solve the equation dydx=1+yx for y(1)=1.
Find the percentage error when y(2) is approximated by the final rounded value found in part (a). Give your answer to two significant figures.
Find the equation of the isocline corresponding to dydx=k, where k≠0, k∈R.
Show that such an isocline can never be a normal to any of the family of curves that satisfy the differential equation.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
attempt to apply Euler’s method (M1)
xn+1=xn+0.25;yn+1=yn+0.25×(1+ynxn)
(A1)(A1)
Note: Award A1 for correct x values, A1 for first three correct y values.
y = 3.3 A1
[4 marks]
METHOD 1
I(x)=e∫−1xdx (M1)
=e−lnx
=1x (A1)
1xdydx−yx2=1x (M1)
ddx(yx)=1x
yx=ln|x|+C A1
y(1)=1⇒C=1 M1
y=xln|x|+x A1
METHOD 2
v=yx M1
dvdx=1xdydx−1x2y (A1)
v+xdvdx=1+v M1
∫1dv=∫1xdx
v=ln|x|+C
yx=ln|x|+C A1
y(1)=1⇒C=1 M1
y=xln|x|+x A1
[6 marks]
y(2)=2ln2+2=3.38629…
percentage error =3.38629…−3.33.38629…×100% (M1)(A1)
= 2.5% A1
[3 marks]
dydx=k⇒1+yx=k A1
y=(k−1)x
[1 mark]
gradient of isocline equals gradient of normal (M1)
k−1=−1k or k(k−1)=−1 A1
k2−k+1=0 A1
Δ=1−4<0 R1
∴ no solution AG
Note: Accept alternative reasons for no solutions.
[4 marks]