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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=y2x2, where x>0.

Solve the differential equation and show that a general solution is x2+y2=cx where c is a positive constant.

[11]
a.

Prove that there are two horizontal tangents to the general solution curve and state their equations, in terms of c.

[5]
b.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

dydx=y2x22xy

let y=vx         M1

dydx=v+xdvdx      (A1)

v+xdvdx=v2x2x22vx2      (M1)

v+xdvdx=v212v(=v212v)      (A1)

Note: Or equivalent attempt at simplification.

xdvdx=v212v(=v212v)      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]

a.

METHOD 1

dydx=y2x22xy

(for horizontal tangents) dydx=0         M1

(y2=x2)y=±x

EITHER

using x2+y2=cx2x2=cx         M1

2x2cx=0x=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=cxx2.

dydx=0x=c2      A1

THEN

tangents at y=c2,y=c2       A1A1

hence there are two tangents     AG

 

METHOD 2

x2+y2=cx

(xc2)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]

b.

Examiners report

[N/A]
a.
[N/A]
b.

Syllabus sections

Topic 5 —Calculus » AHL 5.18—1st order DE’s – Euler method, variables separable, integrating factor, homogeneous DE using sub y=vx
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