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Date May 2018 Marks available 8 Reference code 18M.3.AHL.TZ0.Hca_5
Level Additional Higher Level Paper Paper 3 Time zone Time zone 0
Command term Solve Question number Hca_5 Adapted from N/A

Question

Consider the differential equation  x d y d x y = x p + 1  where  x R , x 0 and  p  is a positive integer,  p > 1 .

Solve the differential equation given that  y = 1 when  x = 1 . Give your answer in the form  y = f ( x ) .

[8]
a.

Show that the x -coordinate(s) of the points on the curve y = f ( x ) where d y d x = 0 satisfy the equation x p 1 = 1 p .

[2]
b.i.

Deduce the set of values for p such that there are two points on the curve y = f ( x ) where d y d x = 0 . Give a reason for your answer.

[2]
b.ii.

Markscheme

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

METHOD 1

d y d x = y x = x p 1 + 1 x     (M1)

integrating factor  = e 1 x d x      M1

 =  e ln x      (A1)

1 x      A1

1 x d y d x y x 2 = x p 2 + 1 x 2      (M1)

d d x ( y x ) = x p 2 + 1 x 2

y x = 1 p 1 x p 1 1 x + C     A1

Note: Condone the absence of C.

y = 1 p 1 x p + C x 1

substituting  x = 1 y = 1 C = 1 p 1     M1 

Note: Award M1 for attempting to find their value of C.

y = 1 p 1 ( x p x ) 1       A1

[8 marks]

 

METHOD 2

put  y = v x so that  d y d x = v + x d v d x     M1(A1)

substituting,       M1 

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

x d v d x = x p 1 + 1 x       M1

d v d x = x p 2 + 1 x 2

v = 1 p 1 x p 1 1 x + C      A1

Note: Condone the absence of C.

y = 1 p 1 x p + C x 1

substituting  x = 1 y = 1 C = 1 p 1     M1 

Note: Award M1 for attempting to find their value of C.

y = 1 p 1 ( x p x ) 1       A1

[8 marks]

a.

METHOD 1

find  d y d x and solve  d y d x = 0 for  x

d y d x = 1 p 1 ( p x p 1 1 )      M1

d y d x = 0 p x p 1 1 = 0      A1

p x p 1 = 1

Note: Award a maximum of M1A0 if a candidate’s answer to part (a) is incorrect.

x p 1 = 1 p      AG

 

METHOD 2

substitute  d y d x = 0  and their y into the differential equation and solve for x

d y d x = 0 ( x p x p 1 ) + 1 = x p + 1      M1

x p x = x p p x p      A1

p x p 1 = 1

Note: Award a maximum of M1A0 if a candidate’s answer to part (a) is incorrect.

x p 1 = 1 p      AG

[2 marks]

 

b.i.

there are two solutions for  x when  p is odd (and  p > 1      A1

if  p 1 is even there are two solutions (to  x p 1 = 1 p )

and if  p 1  is odd there is only one solution (to  x p 1 = 1 p )   R1

Note: Only award the R1 if both cases are considered.

[4 marks]

b.ii.

Examiners report

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

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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