Differential Equations - 15 Multiple Choice + 6 Free Response Questions
Recommended time: 45 minutes
Solution:
1) Separate variables: dy/y = x² dx
2) Integrate both sides: ∫dy/y = ∫x² dx
3) ln|y| = x³/3 + C
4) y = e^(x³/3 + C) = Ce^(x³/3)
5) Use y(0) = 1: 1 = Ce^0 = C, so C = 1
Answer: y = e^(x³/3)
Solution:
1) Start: (x₀, y₀) = (0, 1)
2) Step 1: x₁ = 0.5, y₁ = 1 + 0.5(0 + 1) = 1 + 0.5 = 1.5
3) Step 2: x₂ = 1.0, y₂ = 1.5 + 0.5(0.5 + 1.5) = 1.5 + 1.0 = 2.5
Answer: y(1) ≈ 2.5
Solution:
1) P(t) = P₀e^(kt) where P₀ = 100
2) P(3) = 200 = 100e^(3k), so e^(3k) = 2
3) k = ln(2)/3
4) P(9) = 100e^(9·ln(2)/3) = 100e^(3ln(2)) = 100·2³ = 800
Answer: 800 bacteria
Solution:
1) Separate variables: dy/y = dx/x
2) Integrate both sides: ∫dy/y = ∫dx/x
3) ln|y| = ln|x| + C
4) y = e^(ln|x| + C) = Cx
5) Use y(1) = 3: 3 = C(1), so C = 3
Answer: y = 3x
Solution:
1) M(t) = M₀e^(-kt) where M₀ is initial mass
2) M(10) = 0.25M₀ = M₀e^(-10k), so e^(-10k) = 0.25
3) -10k = ln(0.25) = ln(1/4) = -ln(4), so k = ln(4)/10
4) For half-life: 0.5M₀ = M₀e^(-kt), so t = ln(2)/k = ln(2)/(ln(4)/10) = 10ln(2)/ln(4) = 5 years
Answer: 5 years
Solution:
1) The logistic equation is dy/dt = ky(L - y) where L is carrying capacity
2) Comparing: 0.2y(50 - y) = ky(L - y), so k = 0.2 and L = 50
3) Carrying capacity = 50
4) Long-term behavior: As t → ∞, y → 50 (approaches carrying capacity)
Answer: Carrying capacity = 50, y approaches 50 as t → ∞