Date | May 2019 | Marks available | 11 | Reference code | 19M.3.AHL.TZ0.Hca_5 |
Level | Additional Higher Level | Paper | Paper 3 | Time zone | Time zone 0 |
Command term | Show that and Solve | Question number | Hca_5 | Adapted from | N/A |
Question
Consider the differential equation 2xydydx=y2−x2, where x>0.
Solve the differential equation and show that a general solution is x2+y2=cx where c is a positive constant.
Prove that there are two horizontal tangents to the general solution curve and state their equations, in terms of c.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
dydx=y2−x22xy
let y=vx M1
dydx=v+xdvdx (A1)
v+xdvdx=v2x2−x22vx2 (M1)
v+xdvdx=v2−12v(=v2−12v) (A1)
Note: Or equivalent attempt at simplification.
xdvdx=−v2−12v(=−v2−12v) A1
2v1+v2dvdx=−1x (M1)
∫2v1+v2dv=∫−1xdx (A1)
ln(1+v2)=−lnx+lnc A1A1
Note: Award A1 for LHS and A1 for RHS and a constant.
ln(1+(yx)2)=−lnx+lnc M1
Note: Award M1 for substituting v=yx. May be seen at a later stage.
1+(yx)2=cx A1
Note: Award A1 for any correct equivalent equation without logarithms.
x2+y2=cx AG
[11 marks]
METHOD 1
dydx=y2−x22xy
(for horizontal tangents) dydx=0 M1
(⇒y2=x2)⇒y=±x
EITHER
using x2+y2=cx⇒2x2=cx M1
2x2−cx=0⇒x=c2 A1
Note: Award M1A1 for 2y2=±cy.
OR
using implicit differentiation of x2+y2=cx
2x+2ydydx=c M1
Note: Accept differentiation of y=√cx−x2.
dydx=0⇒x=c2 A1
THEN
tangents at y=c2,y=−c2 A1A1
hence there are two tangents AG
METHOD 2
x2+y2=cx
(x−c2)2+y2=c24 M1A1
this is a circle radius c2 centre (c2,0) A1
hence there are two tangents AG
tangents at y=c2,y=−c2 A1A1
[5 marks]