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Date May 2019 Marks available 5 Reference code 19M.3.AHL.TZ0.Hca_5
Level Additional Higher Level Paper Paper 3 Time zone Time zone 0
Command term Prove and State Question number Hca_5 Adapted from N/A

Question

Consider the differential equation  2 x y d y d x = y 2 x 2 , where  x > 0 .

Solve the differential equation and show that a general solution is  x 2 + y 2 = c x 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.

d y d x = y 2 x 2 2 x y

let  y = v x          M1

d y d x = v + x d v d x       (A1)

v + x d v d x = v 2 x 2 x 2 2 v x 2       (M1)

v + x d v d x = v 2 1 2 v ( = v 2 1 2 v )       (A1)

Note: Or equivalent attempt at simplification.

x d v d x = v 2 1 2 v ( = v 2 1 2 v )       A1

2 v 1 + v 2 d v d x = 1 x       (M1)

2 v 1 + v 2 d v = 1 x d x       (A1)

ln ( 1 + v 2 ) = ln x + ln c        A1A1

Note: Award A1 for LHS and A1 for RHS and a constant.

ln ( 1 + ( y x ) 2 ) = ln x + ln c          M1

Note: Award M1 for substituting  v = y x . May be seen at a later stage.

1 + ( y x ) 2 = c x        A1

Note: Award A1 for any correct equivalent equation without logarithms.

x 2 + y 2 = c x      AG

[11 marks]

a.

METHOD 1

d y d x = y 2 x 2 2 x y

(for horizontal tangents)  d y d x = 0          M1

( y 2 = x 2 ) y = ± x

EITHER

using  x 2 + y 2 = c x 2 x 2 = c x          M1

2 x 2 c x = 0 x = c 2       A1

Note: Award M1A1 for 2 y 2 = ± c y .

OR

using implicit differentiation of  x 2 + y 2 = c x

2 x + 2 y d y d x = c          M1

Note: Accept differentiation of  y = c x x 2 .

d y d x = 0 x = c 2       A1

THEN

tangents at  y = c 2 , y = c 2        A1A1

hence there are two tangents     AG

 

METHOD 2

x 2 + y 2 = c x

( x c 2 ) 2 + y 2 = c 2 4        M1A1

this is a circle radius  c 2 centre  ( c 2 , 0 )        A1

hence there are two tangents     AG

tangents at  y = c 2 , y = c 2        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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Topic 5 —Calculus

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