Q.13 Let 𝑦(𝑡) be a bacterial population whose growth is given by
dy/dt=𝜆(𝑦 + 2)
where 𝜆 is the growth rate constant. If 𝑦(0) = 1 and 𝑦(1) = 4, then the value of
𝜆 is
(A) ln 2
(B) ln 3
(C) ln 4
(D) ln 6
This separable differential equation models bacterial growth proportional to (y + 2). Solving with initial conditions y(0) = 1 and y(1) = 4 yields λ = ln 2.
Solving the DE Step-by-Step
Step 2: Integrate both sides: ∫ dy/(y + 2) = λ ∫ dt
giving ln|y + 2| = λt + C
Step 3: Exponentiate: y + 2 = C’ eλt, so y(t) = C’ eλt – 2
Step 4: Apply y(0) = 1: 1 = C’ – 2, thus C’ = 3
and y(t) = 3 eλt – 2
Step 5: Apply y(1) = 4: 4 = 3 eλ – 2
so 6 = 3 eλ, eλ = 2, λ = ln 2
Why Option (A) ln 2 is Correct
λ = ln 2 satisfies both conditions exactly, as eln 2 = 2 leads to y(1) = 3(2) – 2 = 4.
This matches the population tripling effectively from the shifted baseline (y + 2 from 3 to 6).
Explanation of Incorrect Options
- (B) ln 3: eln 3 ≈ 3, y(1) = 3(3) – 2 = 7 > 4, overestimates growth.
- (C) ln 4: eln 4 = 4, y(1) = 3(4) – 2 = 10 >> 4, far too rapid.
- (D) ln 6: eln 6 ≈ 6, y(1) = 3(6) – 2 = 16, excessively high.
Options Comparison Table
| Option | λ Value | Predicted y(1) | Matches y(1)=4? |
|---|---|---|---|
| (A) ln 2 | ≈0.693 | 4 | Yes |
| (B) ln 3 | ≈1.099 | 7 | No |
| (C) ln 4 | ≈1.386 | 10 | No |
| (D) ln 6 | ≈1.792 | 16 | No |
