 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ib/paper-4/12b/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

t \geq 0

f (s) = L {f} (s) = \int_{0}^{\infty} e^{- s t} f (t) d t,

for values of

s

L {f^{'}} (s) = s f (s) - f (0) .

Write

N_{i} (s) = L {N_{i}} (s)

. Transforming the first two equations and using the initial data gives

N_{1} (s) = \frac{N}{s + λ _{1}}, N_{2} (s) = \frac{N λ _{1}}{( s + λ _{1} ) ( s + λ _{2} )} .

The third equation then gives

(s + λ_{3}) N_{3} (s) - n = λ_{2} N_{2} (s),

and hence

N_{3} (s) = \frac{n}{s + λ _{3}} + \frac{N λ _{1} λ _{2}}{( s + λ _{1} ) ( s + λ _{2} ) ( s + λ _{3} )} .

N_{3} (t) = n e^{- λ_{3} t} + N λ_{1} λ_{2} [\frac{e ^{- λ_{1} t}}{( λ _{2} - λ _{1} ) ( λ _{3} - λ _{1} )} + \frac{e ^{- λ_{2} t}}{( λ _{1} - λ _{2} ) ( λ _{3} - λ _{2} )} + \frac{e ^{- λ_{3} t}}{( λ _{1} - λ _{3} ) ( λ _{2} - λ _{3} )}] .

Now let

λ_{1} = λ_{2} = λ

and put

Δ = λ_{3} - λ > 0

. The transformed contribution originating from the initial

N_{1}

\frac{N λ ^{2}}{( s + λ ) ^{2} ( s + λ _{3} )} .

Equivalently, solve the middle equation first to obtain

N_{2} (t) = N λ t e^{- λ t},

N_{3} (t) = n e^{- λ_{3} t} + N λ^{2} e^{- λ_{3} t} \int_{0}^{t} u e^{Δ u} d u .

Evaluating the elementary integral gives

N_{3} (t) = n e^{- λ_{3} t} + \frac{N λ ^{2}}{( λ _{3} - λ ) ^{2}} [e^{- λ t} ((λ_{3} - λ) t - 1) + e^{- λ_{3} t}] .

 New to topics? Read the docs here!