A mass–spring–damper is described by y″ + 5.6y′ + 16y = 0. Compare it with y″ + 2ζωny′ + ωn²y = 0 and find the damping ratio ζ.
Differential equations worksheet with answers
1.Easy
y″ + 5.6y′ + 16y = 0
ζ =
2.Easy
For the equation y″ + 4y′ + 3y = 0, find the root r of the characteristic equation that is closest to zero. It sets how slowly the solution dies out.
y″ + 4y′ + 3y = 0
r =
3.Medium
Solve y″ + 4y′ + 4y = 0 with y(0) = 1 and y′(0) = 2. Find y(0.5).
y″ + 4y′ + 4y = 0y(0) = 1y′(0) = 2
y(0.5) =
4.Medium
A forged steel part at 300 °C is left to cool in a workshop at 28 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.05 min⁻¹, find its temperature T after 31 min.
T0 = 300 °CTa = 28 °Ck = 0.05 min⁻¹t = 31 min
T = °C
5.Hard
A system obeys y″ + 4y′ + 20y = 40 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find the peak value ymax of its step response.
y″ + 4y′ + 20y = 40y(0) = 0y′(0) = 0
ymax =
6.Hard
Solve the initial value problem y′ + y/x = 2x for x > 0, with y(1) = 5. Find y(3).
y′ + y/x = 2xy(1) = 5
y(3) =
Answer key
- 1. ζ = 0.7
- 2. r = −1
- 3. y(0.5) = 1.104
- 4. T = 85.73 °C
- 5. ymax = 2.416
- 6. y(3) = 7.444
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