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Date November 2021 Marks available 3 Reference code 21N.3.AHL.TZ0.2
Level Additional Higher Level Paper Paper 3 Time zone Time zone 0
Command term Show that and Hence Question number 2 Adapted from N/A

Question

In this question you will be exploring the strategies required to solve a system of linear differential equations.

 

Consider the system of linear differential equations of the form:

dxdt=x-y  and  dydt=ax+y,

where x, y, t+ and a is a parameter.

First consider the case where a=0.

Now consider the case where a=-1.

Now consider the case where a=-4.

From previous cases, we might conjecture that a solution to this differential equation is y=Feλt, λ and F is a constant.

By solving the differential equation dydt=y, show that y=Aet where A is a constant.

[3]
a.i.

Show that dxdt-x=-Aet.

[1]
a.ii.

Solve the differential equation in part (a)(ii) to find x as a function of t.

[4]
a.iii.

By differentiating dydt=-x+y with respect to t, show that d2ydt2=2dydt.

[3]
b.i.

By substituting Y=dydt, show that Y=Be2t where B is a constant.

[3]
b.ii.

Hence find y as a function of t.

[2]
b.iii.

Hence show that x=-B2e2t+C, where C is a constant.

[3]
b.iv.

Show that d2ydt2-2dydt-3y=0.

[3]
c.i.

Find the two values for λ that satisfy d2ydt2-2dydt-3y=0.

[4]
c.ii.

Let the two values found in part (c)(ii) be λ1 and λ2.

Verify that y=Feλ1t+Geλ2t is a solution to the differential equation in (c)(i),where G is a constant.

[4]
c.iii.

Markscheme

METHOD 1

dydt=y

dyy=dt               (M1)

lny=t+c  OR  lny=t+c             A1A1


Note: Award A1 for lny and A1 for t and c.


y=Aet             AG

 

METHOD 2

rearranging to dydt-y=0 AND multiplying by integrating factor e-t               M1

ye-t=A             A1A1

y=Aet             AG

 

[3 marks]

a.i.

substituting y=Aet into differential equation in x               M1

dxdt=x-Aet

dxdt-x=-Aet             AG

 

[1 mark]

a.ii.

integrating factor (IF) is e-1dt               (M1)

=e-t               (A1)

e-tdxdt-xe-t=-A

xe-t=-At+D               (A1)

x=-At+Det               A1


Note: The first constant must be A, and the second can be any constant for the final A1 to be awarded. Accept a change of constant applied at the end.

 

[4 marks]

a.iii.

d2ydt2=-dxdt+dydt               A1


EITHER

=-x+y+dydt               (M1)

=dydt+dydt               A1


OR

=-x+y+-x+y               (M1)

=2-x+y               A1


THEN

=2dydt               AG


[3 marks]

b.i.

dYdt=2Y               A1

dYY=2dt               M1

lnY=2t+c  OR  lnY=2t+c               A1

Y=Be2t               AG

 

[3 marks]

b.ii.

dydt=Be2t

y=Be2tdt              M1

y=B2e2t+C              A1


Note:
The first constant must be B, and the second can be any constant for the final A1 to be awarded. Accept a change of constant applied at the end.

 

[2 marks]

b.iii.

METHOD 1

substituting dydt=Be2t and their (iii) into dydt=-x+y              M1(M1)

Be2t=-x+B2e2t+C              A1

x=-B2e2t+C              AG

Note: Follow through from incorrect part (iii) cannot be awarded if it does not lead to the AG.


METHOD 2

dxdt=x-B2e2t-C

dxdt-x=-B2e2t-C

dxe-tdt=-B2et-Ce-t              M1

xe-t=-B2et-Ce-tdt

xe-t=-B2et-Ce-t+D              A1

x=-B2e2t+C+Det

dydt=-x+yBe2t=B2e2t-C-Det+B2e2t+CD=0              M1

x=-B2e2t+C              AG

 

[3 marks]

b.iv.

dydt=-4x+y

d2ydt2=-4dxdt+dydt seen anywhere              M1

 

METHOD 1

d2ydt2=-4x-y+dydt

attempt to eliminate x              M1

=-414y-dydt-y+dydt

=2dydt+3y              A1

d2ydt2-2dydt-3y=0              AG

 

METHOD 2

rewriting LHS in terms of x and y              M1

d2ydt2-2dydt-3y=-8x+5y-2-4x+y-3y              A1

=0              AG

 

[3 marks]

c.i.

dydt=Fλeλt, d2ydt2=Fλ2eλt               (A1)

Fλ2eλt-2Fλeλt-3Feλt=0               (M1)

λ2-2λ-3=0  (since eλt0)              A1

λ1 and λ2 are 3 and -1 (either order)              A1

 

[4 marks]

c.ii.

METHOD 1

y=Fe3t+Ge-t

dydt=3Fe3t-Ge-t, d2ydt2=9Fe3t-Ge-t                      (A1)(A1)

d2ydt2-2dydt-3y=9Fe3t+Ge-t-23Fe3t-Ge-t-3Fe3t-Ge-t              M1

=9Fe3t+Ge-t-6Fe3t+2Ge-t-3Fe3t-3Ge-t              A1

=0              AG

 

METHOD 2

y=Feλ1t+Geλ2t

dydt=Fλ1eλ1t+Gλ2eλ2t, d2ydt2=Fλ12eλ1t+Gλ22eλ2t                      (A1)(A1)

d2ydt2-2dydt-3y=Fλ12eλ1t+Gλ22eλ2t-2Fλ1eλ1t+Gλ2eλ2t-3Feλ1t+Geλ2t              M1

=Feλ1tλ2-2λ-3+Geλ2tλ2-2λ-3              A1

=0              AG

 

[4 marks]

c.iii.

Examiners report

[N/A]
a.i.
[N/A]
a.ii.
[N/A]
a.iii.
[N/A]
b.i.
[N/A]
b.ii.
[N/A]
b.iii.
[N/A]
b.iv.
[N/A]
c.i.
[N/A]
c.ii.
[N/A]
c.iii.

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