IB Maths AI HL 5.18 Second-order differential equations Question Bank

Practise IB Mathematics HL 5.18 by applying second-order differential equations methods to exam-style questions.

[Maximum number: 10]

An environmental scientist is asked by a river authority to model the effect of a leak from a power plant on the mercury levels in a local river. The variable x measures the concentration of mercury in micrograms per litre.
The situation is modelled using the second order differential equation

\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+3 \frac{\mathrm{~d} x}{\mathrm{~d} t}+2 x=0

where $t \geq 0$ is the time measured in days since the leak started. It is known that when t=0, x=0 and $\frac{\mathrm{d} x}{\mathrm{~d} t}=1$ .

(a)

Show that the system of coupled first order equations:

$\frac{\mathrm{d}x}{\mathrm{d}t}=y$ $\frac{\mathrm{d}y}{\mathrm{d}t}=-2x-3y$

can be written as the given second order differential equation.

[ 2 ]

differentiating first equation.
$\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}=\frac{\mathrm{d} y}{\mathrm{~d} t}$
substituting in for $\frac{\mathrm{d} y}{\mathrm{~d} t}$ $=-2 x-3 y=-2 x-3 \frac{\mathrm{~d} x}{\mathrm{~d} t}$
therefore $\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+3 \frac{\mathrm{~d} x}{\mathrm{~d} t}+2 x=0$

Note: The AG line must be seen to award the final M1 mark.

(b)

Hence find the exact solution of the second order differential equation.

[ 5 ]

EITHER
The general solution is $x=A\mathrm{e}^{-t}+B\mathrm{e}^{-2t}$ . M1
Note: Must have constants, but condone sign error for the M1.
Then $\frac{\mathrm{d}x}{\mathrm{d}t}=-A\mathrm{e}^{-t}-2B\mathrm{e}^{-2t}$ .
OR
Attempt to find eigenvectors: respective eigenvectors are $\binom1{-1}$ and $\binom1{-2}$ .
$\binom{x}{y}=A\mathrm{e}^{-t}\binom1{-1}+B\mathrm{e}^{-2t}\binom1{-2}$
THEN
Initial conditions: 0=A+B, 1=-A-2B.
This gives A=1, B=-1, so $x=\mathrm{e}^{-t}-\mathrm{e}^{-2t}$ .

(c)

If the mercury levels are greater than 0.1 micrograms per litre, fishing in the river is considered unsafe and is stopped.
Use the model to calculate the total amount of time when fishing should be stopped.

[ 3 ]

intersecting graph with y=0.1

Figure for Answer to Question (c) — IB Maths AI HL

so the time fishing is stopped between $2.1830 \ldots$ and $0.11957 \ldots$
=2.06 ( $343 \ldots$ ) days

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AHL 5.18 (HL)—Second-order differential equations question 1

Question (a)

Question (b)

Question (c)