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Date May 2016 Marks available 5 Reference code 16M.2.hl.TZ0.4
Level HL only Paper 2 Time zone TZ0
Command term Hence and Sketch Question number 4 Adapted from N/A

Question

Consider the differential equation dydx=xy, where y0.

Find the general solution of the differential equation, expressing your answer in the form f(x, y)=c, where c is a constant.

[3]
a.

(i)     Hence find the particular solution passing through the points (1,±2).

(ii)     Sketch the graph of your solution and name the type of curve represented.

[5]
b.

(i)     Write down the particular solution passing through the points (1, ±1).

(ii)     Give a geometrical interpretation of this solution in relation to part (b).

[3]
c.

(i)     Find the general solution of the differential equation dydx=xy+yx, where xy0.

(ii)     Find the particular solution passing through the point (1, 2).

(iii)     Sketch the particular solution.

(iv)     The graph of the solution only contains points with |x|>a.

Find the exact value of a, a>0.

[12]
d.

Markscheme

attempt to separate the variables     M1

ydydxdx=xdx    A1

 

Note:     Accept ydy=xdx.

 

obtain 12y2=12x2+ constant (y2x2=c)     A1

[3 marks]

a.

(i)     substitute the coordinates for both points     M1

(±2)212=1

obtain y2x2=1 or equivalent     A1

(ii)     M16/5/FURMA/HP2/ENG/TZ0/04.b.ii/M     A1A1

 

Note:     A1 for general shape including two branches and symmetry;

A1 for values of the intercepts.

 

(rectangular) hyperbola     A1

[5 marks]

b.

(i)     y2x2=0     A1

(ii)     the two straight lines y=±x     A1

form the asymptotes to the hyperbola found above, or equivalent     A1

[3 marks]

c.

(i)     the equation is homogeneous, so attempt to substitute y=vx     M1

as a first step write dydx=xdvdx+v     (A1)

then xdvdx+v=1v+v     A1

attempt to solve the resulting separable equation     M1

vdv=1xdx    A1

obtain 12v2=ln|x|+ constanty2=2x2ln|x|+cx2     A1

(ii)     substituting the coordinates     (M1)

obtain c=2y2=2x2ln|x|+2x2     A1

(iii)     M16/5/FURMA/HP2/ENG/TZ0/04.d.iii/M     A1

(iv)     since y2>0 and x20     R1

ln|x|>1|x|>e1    A1

a=e1    A1

 

Note:     The R1 may be awarded for a correct reason leading to subsequent correct work.

 

[12 marks]

d.

Examiners report

Although (a), (b) and (c) were generally well done, it was rare to see a completely satisfactory geometrical answer to part (c)(ii). A few candidates solved the differential equation as a homogeneous equation.

a.

Although (a), (b) and (c) were generally well done, it was rare to see a completely satisfactory geometrical answer to part (c)(ii). A few candidates solved the differential equation as a homogeneous equation.

b.

Although (a), (b) and (c) were generally well done, it was rare to see a completely satisfactory geometrical answer to part (c)(ii). A few candidates solved the differential equation as a homogeneous equation.

c.

For part (d) most candidates used the correct solution method for a homogeneous differential equation. A few found the algebra hard going in finding the particular solution. Most approaches to the final part were unsatisfactory, with a lack of proper consideration of the inequalities in the question.

d.

Syllabus sections

Topic 5 - Calculus » 5.5 » First-order differential equations.
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