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IB Mathematics: Applications and Interpretation HL · AHL 9: Differential Equations · AHL 9.15
Mixed practice: stability of second order equations Problem set A One question at a time
Try each question on paper first. Open the working a line at a time, then say whether you had it.
Skills
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Write each equation as a system of coupled first order equations, using y = dx/dt, and give the matrix of each system.
(a) \(\frac{d^2x}{dt^2} + 9\frac{dx}{dt} + 14x = 0\)
- Let y = dx/dt
- \(\frac{dx}{dt} = y,\quad \frac{dy}{dt} = -9y - 14x\)
- Matrix \(\begin{pmatrix} 0 & 1 \\ -14 & -9 \end{pmatrix}\)
(b) \(\frac{d^2x}{dt^2} - 2\frac{dx}{dt} + 10x = 0\)
- Let y = dx/dt
- \(\frac{dx}{dt} = y,\quad \frac{dy}{dt} = 2y - 10x\)
- Matrix \(\begin{pmatrix} 0 & 1 \\ -10 & 2 \end{pmatrix}\)
(c) \(\frac{d^2x}{dt^2} = 7x - 3\frac{dx}{dt}\)
- Replace dx/dt by y and d²x/dt² by dy/dt
- \(\frac{dy}{dt} = 7x - 3y\)
- Matrix \(\begin{pmatrix} 0 & 1 \\ 7 & -3 \end{pmatrix}\)
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Decide whether the solution of each equation is stable. Give the eigenvalues.
(a) \(\frac{d^2x}{dt^2} + 8\frac{dx}{dt} + 7x = 0\)
- Characteristic equation \(\lambda^2 + 8\lambda + 7 = 0\)
- λ = −1 and −7
- Stable: both eigenvalues are negative
(b) \(\frac{d^2x}{dt^2} + 3\frac{dx}{dt} - 10x = 0\)
- Characteristic equation \(\lambda^2 + 3\lambda - 10 = 0\)
- λ = 2 and −5
- Unstable: one eigenvalue is positive (a saddle)
(c) \(\frac{d^2x}{dt^2} - 7\frac{dx}{dt} + 12x = 0\)
- Characteristic equation \(\lambda^2 - 7\lambda + 12 = 0\)
- λ = 4 and 3
- Unstable: both eigenvalues are positive
(d) \(\frac{d^2x}{dt^2} - 49x = 0\)
- Characteristic equation \(\lambda^2 - 49 = 0\)
- λ = 7 and −7
- Unstable: one eigenvalue is positive (a saddle)
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Write each equation as a system and decide whether the solution is stable. Give the eigenvalues.
(a) \(\frac{d^2x}{dt^2} + 4\frac{dx}{dt} + 20x = 0\)
- Characteristic equation \(\lambda^2 + 4\lambda + 20 = 0\)
- λ = −2 ± 4i
- Stable: the real part −2 is negative
(b) \(\frac{d^2x}{dt^2} - 2\frac{dx}{dt} + 5x = 0\)
- Characteristic equation \(\lambda^2 - 2\lambda + 5 = 0\)
- λ = 1 ± 2i
- Unstable: the real part 1 is positive
(c) \(\frac{d^2x}{dt^2} + 16x = 0\)
- Characteristic equation \(\lambda^2 + 16 = 0\)
- λ = ±4i
- Not stable: the real part is 0: x keeps oscillating and does not tend to 0
In context
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The vertical displacement x metres of a car body on its suspension, t seconds after a bump, satisfies \(\frac{d^2x}{dt^2} + c\frac{dx}{dt} + 36x = 0\), where c is a positive constant.
(a) For c = 3, write the equation as a system and give its matrix.
- \(\frac{dx}{dt} = y,\quad \frac{dy}{dt} = -3y - 36x\)
- Matrix \(\begin{pmatrix} 0 & 1 \\ -36 & -3 \end{pmatrix}\)
(b) Find the eigenvalues for c = 3 and say whether the motion dies away.
- Characteristic equation \(\lambda^2 + 3\lambda + 36 = 0\)
- λ = −1.5 ± 5.81i
- Stable: the real part −1.5 is negative
(c) Find the eigenvalues for c = 15 and say whether the motion dies away.
- Characteristic equation \(\lambda^2 + 15\lambda + 36 = 0\)
- λ = −3 and −12
- Stable: both eigenvalues are negative
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The temperature x °C above target in a faulty heating system, t minutes after it is switched on, satisfies \(\frac{d^2x}{dt^2} - 0.6\frac{dx}{dt} + 5x = 0\).
(a) Find the eigenvalues of the system.
- Matrix \(\begin{pmatrix} 0 & 1 \\ -5 & 0.6 \end{pmatrix}\)
- Characteristic equation \(\lambda^2 - 0.6\lambda + 5 = 0\)
- λ = 0.3 ± 2.22i
(b) Is the solution stable? Say what this means for the temperature.
- The real part is 0.3, which is positive
- This means that the temperature swings above and below the target by larger and larger amounts, so the system does not settle
Stretch
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Consider \(\frac{d^2x}{dt^2} + k\frac{dx}{dt} + 9x = 0\).
(a) Find the eigenvalues, and decide stability, when k = 2.
- Characteristic equation \(\lambda^2 + 2\lambda + 9 = 0\)
- λ = −1 ± 2.83i
- Stable: the real part −1 is negative
(b) Find the eigenvalues, and decide stability, when k = 10.
- Characteristic equation \(\lambda^2 + 10\lambda + 9 = 0\)
- λ = −1 and −9
- Stable: both eigenvalues are negative
(c) Without solving, say what happens to the stability when k = 0 and when k is negative. Give a reason.
- When k = 0: λ² + 9 = 0, so λ = ±3i, real part 0: x does not tend to 0
- When k < 0: the sum of the eigenvalues is −k > 0, so the real part, or at least one real eigenvalue, is positive: unstable
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Find all values of b for which the solution of \(\frac{d^2x}{dt^2} + 4\frac{dx}{dt} + bx = 0\) is stable. Give a reason.
- Characteristic equation λ² + 4λ + b = 0, so λ = −2 ± √(4 − b)
- If b > 4 the roots are complex with real part −2: stable
- If 0 < b ≤ 4 the roots are real and √(4 − b) < 2, so both are negative: stable
- If b = 0 one root is 0, and if b < 0 one root is positive: not stable. So stable exactly when b > 0
jmaths.xyz · AI HL · AHL 9.15 Stability of second order equations