1.Easy
A mass–spring–damper is described by y″ + 1.6y′ + 16y = 0. Compare it with y″ + 2ζωny′ + ωn²y = 0 and find the damping ratio ζ.
y″ + 1.6y′ + 16y = 0
ζ =
Differential equations worksheet with answers
2.Easy
A quantity grows according to dy/dt = k·y with k > 0. It is 500 at t = 0 and 950 at t = 2 h. Find the rate constant k in h⁻¹.
y(0) = 500y(2) = 950
k = h⁻¹
3.Medium
A forged steel part at 400 °C is left to cool in a workshop at 30 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.02 min⁻¹, find its temperature T after 50 min.
T0 = 400 °CTa = 30 °Ck = 0.02 min⁻¹t = 50 min
T = °C
4.Medium
A can of soda at 4 °C is taken out of the fridge into a room at 28 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.09 min⁻¹, find its temperature T after 11 min.
T0 = 4 °CTa = 28 °Ck = 0.09 min⁻¹t = 11 min
T = °C
5.Hard
A tank holds 100 L of water in which 10 kg of salt is dissolved. Brine with 0.05 kg of salt per litre flows in at 5 L/min, and the well-mixed solution flows out at the same rate. After how many minutes t does the tank contain 7 kg of salt?
V = 100 Lm0 = 10 kgc = 0.05 kg/Lr = 5 L/min
t = min
6.Hard
A system obeys y″ + 4y′ + 29y = 58 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find the peak value ymax of its step response.
y″ + 4y′ + 29y = 58y(0) = 0y′(0) = 0
ymax =
Answer key
- 1. ζ = 0.2
- 2. k = 0.3209 h⁻¹
- 3. T = 166.1 °C
- 4. T = 19.08 °C
- 5. t = 18.33 min
- 6. ymax = 2.569
Full step-by-step solutions: practise this topic on examproblems.com.