Date | November 2021 | Marks available | 7 | Reference code | 21N.1.AHL.TZ0.8 |
Level | Additional Higher Level | Paper | Paper 1 (without calculator) | Time zone | Time zone 0 |
Command term | Solve | Question number | 8 | Adapted from | N/A |
Question
Solve the differential equation dydx=ln 2xx2-2yx, x>0, given that y=4 at x=12.
Give your answer in the form y=f(x).
Markscheme
dydx+2yx=ln 2xx2 (M1)
attempt to find integrating factor (M1)
(e∫2xdx=e2 ln x)=x2 (A1)
x2dydx+2xy=ln 2x
ddx(x2y)=ln 2x
x2y=∫ln 2x dx
attempt to use integration by parts (M1)
x2y=x ln 2x-x(+c) A1
y=ln 2xx-1x+cx2
substituting x=12, y=4 into an integrated equation involving c M1
4=0-2+4c
⇒c=32
y=ln 2xx-1x+32x2 A1
[7 marks]
Examiners report
[N/A]