IB Questionbank

User interface language: English | Español

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:

$\frac{d x}{d t} = x - y$ and $\frac{d y}{d t} = a x + y$ ,

where $x, y, t \in ℝ^{+}$ 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 $y = F e^{λ t}$ , $λ \in ℝ$ and $F$ is a constant.

By solving the differential equation $\frac{d y}{d t} = y$ , show that $y = A e^{t}$ where $A$ is a constant.

[3]

a.i.

Show that $\frac{d x}{d t} - x = - A e^{t}$ .

[1]

a.ii.

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

[4]

a.iii.

By differentiating $\frac{d y}{d t} = - x + y$ with respect to $t$ , show that $\frac{d^{2} y}{d t^{2}} = 2 \frac{d y}{d t}$ .

[3]

b.i.

By substituting $Y = \frac{d y}{d t}$ , show that $Y = B e^{2 t}$ where $B$ is a constant.

[3]

b.ii.

Hence find $y$ as a function of $t$ .

[2]

b.iii.

Hence show that $x = - \frac{B}{2} e^{2 t} + C$ , where $C$ is a constant.

[3]

b.iv.

Show that $\frac{d^{2} y}{d t^{2}} - 2 \frac{d y}{d t} - 3 y = 0$ .

[3]

c.i.

Find the two values for $λ$ that satisfy $\frac{d^{2} y}{d t^{2}} - 2 \frac{d y}{d t} - 3 y = 0$ .

[4]

c.ii.

Let the two values found in part (c)(ii) be $λ_{1}$ and $λ_{2}$ .

Verify that $y = F e^{λ_{1} t} + G e^{λ_{2} t}$ is a solution to the differential equation in (c)(i),where $G$ is a constant.

[4]

c.iii.

Markscheme

METHOD 1

$\frac{d y}{d t} = y$

$\int \frac{d y}{y} = \int d t$ (M1)

$\ln y = t + c$ OR $\ln |y| = t + c$ A1A1

Note: Award A1 for $\ln y$ and A1 for $t$ and $c$ .

$y = A e^{t}$ AG

METHOD 2

rearranging to $\frac{d y}{d t} - y = 0$ AND multiplying by integrating factor $e^{- t}$ M1

$y e^{- t} = A$ A1A1

$y = A e^{t}$ AG

[3 marks]

a.i.

substituting $y = A e^{t}$ into differential equation in $x$ M1

$\frac{d x}{d t} = x - A e^{t}$

$\frac{d x}{d t} - x = - A e^{t}$ AG

[1 mark]

a.ii.

integrating factor (IF) is $e^{\int - 1 d t}$ (M1)

$= e^{- t}$ (A1)

$e^{- t} \frac{d x}{d t} - x e^{- t} = - A$

$x e^{- t} = - A t + D$ (A1)

$x = (- A t + D) e^{t}$ 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.

$\frac{d^{2} y}{d t^{2}} = - \frac{d x}{d t} + \frac{d y}{d t}$ A1

EITHER

$= - x + y + \frac{d y}{d t}$ (M1)

$= \frac{d y}{d t} + \frac{d y}{d t}$ A1

$= - x + y + (- x + y)$ (M1)

$= 2 (- x + y)$ A1

THEN

$= 2 \frac{d y}{d t}$ AG

[3 marks]

b.i.

$\frac{d Y}{d t} = 2 Y$ A1

$\int \frac{d Y}{Y} = \int 2 d t$ M1

$\ln |Y| = 2 t + c$ OR $\ln Y = 2 t + c$ A1

$Y = B e^{2 t}$ AG

[3 marks]

b.ii.

$\frac{d y}{d t} = B e^{2 t}$

$y = \int B e^{2 t} d t$ M1

$y = \frac{B}{2} e^{2 t} + C$ 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 $\frac{d y}{d t} = B e^{2 t}$ and their (iii) into $\frac{d y}{d t} = - x + y$ M1(M1)

$B e^{2 t} = - x + \frac{B}{2} e^{2 t} + C$ A1

$x = - \frac{B}{2} e^{2 t} + C$ AG

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

METHOD 2

$\frac{d x}{d t} = x - \frac{B}{2} e^{2 t} - C$

$\frac{d x}{d t} - x = - \frac{B}{2} e^{2 t} - C$

$\frac{d (x e^{- t})}{d t} = - \frac{B}{2} e^{t} - C e^{- t}$ M1

$x e^{- t} = \int - \frac{B}{2} e^{t} - C e^{- t} d t$

$x e^{- t} = - \frac{B}{2} e^{t} - C e^{- t} + D$ A1

$x = - \frac{B}{2} e^{2 t} + C + D e^{t}$

$\frac{d y}{d t} = - x + y \Rightarrow B e^{2 t} = \frac{B}{2} e^{2 t} - C - D e^{t} + \frac{B}{2} e^{2 t} + C \Rightarrow D = 0$ M1

$x = - \frac{B}{2} e^{2 t} + C$ AG

[3 marks]

b.iv.

$\frac{d y}{d t} = - 4 x + y$

$\frac{d^{2} y}{d t^{2}} = - 4 \frac{d x}{d t} + \frac{d y}{d t}$ seen anywhere M1

METHOD 1

$\frac{d^{2} y}{d t^{2}} = - 4 (x - y) + \frac{d y}{d t}$

attempt to eliminate $x$ M1

$= - 4 (\frac{1}{4} (y - \frac{d y}{d t}) - y) + \frac{d y}{d t}$

$= 2 \frac{d y}{d t} + 3 y$ A1

$\frac{d^{2} y}{d t^{2}} - 2 \frac{d y}{d t} - 3 y = 0$ AG

METHOD 2

rewriting LHS in terms of $x$ and $y$ M1

$\frac{d^{2} y}{d t^{2}} - 2 \frac{d y}{d t} - 3 y = (- 8 x + 5 y) - 2 (- 4 x + y) - 3 y$ A1

$= 0$ AG

[3 marks]

c.i.

$\frac{d y}{d t} = F λ e^{λ t}, \frac{d^{2} y}{d t^{2}} = F λ^{2} e^{λ t}$ (A1)

$F λ^{2} e^{λ t} - 2 F λ e^{λ t} - 3 F e^{λ t} = 0$ (M1)

$λ^{2} - 2 λ - 3 = 0$ (since $e^{λ t} \neq 0$ ) A1

$λ_{1}$ and $λ_{2}$ are $3$ and $- 1$ (either order) A1

[4 marks]

c.ii.

METHOD 1

$y = F e^{3 t} + G e^{- t}$

$\frac{d y}{d t} = 3 F e^{3 t} - G e^{- t}, \frac{d^{2} y}{d t^{2}} = 9 F e^{3 t} - G e^{- t}$ (A1)(A1)

$\frac{d^{2} y}{d t^{2}} - 2 \frac{d y}{d t} - 3 y = 9 F e^{3 t} + G e^{- t} - 2 (3 F e^{3 t} - G e^{- t}) - 3 (F e^{3 t} - G e^{- t})$ M1

$= 9 F e^{3 t} + G e^{- t} - 6 F e^{3 t} + 2 G e^{- t} - 3 F e^{3 t} - 3 G e^{- t}$ A1

$= 0$ AG

METHOD 2

$y = F e^{λ_{1} t} + G e^{λ_{2} t}$

$\frac{d y}{d t} = F λ_{1} e^{λ_{1} t} + G λ_{2} e^{λ_{2} t}, \frac{d^{2} y}{d t^{2}} = F {λ_{1}}^{2} e^{λ_{1} t} + G {λ_{2}}^{2} e^{λ_{2} t}$ (A1)(A1)

$\frac{d^{2} y}{d t^{2}} - 2 \frac{d y}{d t} - 3 y = F {λ_{1}}^{2} e^{λ_{1} t} + G {λ_{2}}^{2} e^{λ_{2} t} - 2 (F λ_{1} e^{λ_{1} t} + G λ_{2} e^{λ_{2} t}) - 3 (F e^{λ_{1} t} + G e^{λ_{2} t})$ M1