Exercise from Calculus 'Separation of Variables'

Question

At any time t, the rate of growth of the population N of deer in a state park is proportional to the product of N and L-N, where L=500 is the maximum number of deer the park can sustain. When t=0, N=100, and when t=4, N=200. Write N as a function of t.

Answer 1

Don't fall into the trap of conceptualizing. They verbatim told you:

1. N' = kN(L-N) (that's the definition of "proportional to")
2. L=500
3. N(0) = 100
4. N(4) = 200

Rewrite as dN/[N(L-N)] = k dt and integrate both sides and you're done.

1. You may want to use partial fractions.
2. You may find a definite integral (0 to 4) is faster than an indefinite one

Answer 2

Given:

Rate of growth of population = dN/dt = k₀ [N(L - N)]

L = 500 , Maximum population

N(0) = 100

N(4) = 200

dN/dt = k₀ [N(L - N)] ==> dN / [N(L - N)] = k₀dt

1 / N(L - N) = [ A/N + B/(L - N) ] ==> [A(L - N) + BN] = 1 ==> (B - A)N = 0, AL = 1 ==> A = B = 1/L

(1/L)(1/N + 1/(L - N))dN = k₀dt

dN/N + dN/(L - N) = (k₀L)dt

ln | N | + ln | L - N | = ln | N(L - N) | = k₀Lt + C

N(L - N) = k₁eᵃ⁰ᵀ , a₀ = k₀L

LN - N² = k₁eᵃ⁰ᵀ

N² - LN + L²/4 = -k₁eᵃ⁰ᵀ + L²/4

(N - L/2)² = L²/4 - k₁eᵃ⁰ᵀ

N - L/2 = ±√(L²/4 - k₁eᵃ⁰ᵀ)

N(t) = ±√(L²/4 - k₁eᵃ⁰ᵀ) + L/2

N(0) = 100 = ±√(L²/4 - k₁) + L/2 ==> 100 - 250 = ±√(L²/4 - k₁) ==> L²/4 - k₁ = 150² ==> k₁ = (L/2)² - 150² = 250² - 150² = 10²(25² - 15²) = 10²(625 - 225) = 10²(400) ==> k₁ = 4 x 10⁴

(NB) There is another conclusion here: A negative sign before the √() in N(t) = L/2 - √(L²/4 - k₁eᵃ⁰ᵀ)... which we get from above: 100 - 250 = -150 = √(L²/4 - k₁), so we wanted the negative root: -√()

N(4) = 200 = 250 - √(L²/4 - k₁e⁴ᵃ⁰) ==> -√(L²/4 - k₁e⁴ᵃ⁰) = -50 ==> L²/4 - k₁e⁴ᵃ⁰ = 50² ==> k₁e⁴ᵃ⁰ = (L/2)² - 50² = 250² - 50² = 10²(25² - 5²) = 10²(600) ==> [4 x 10⁴]e⁴ᵃ⁰ = 6 x 10⁴

e⁴ᵃ⁰ = 6 x 10⁴ / 4 x 10⁴ = 6/4 = 3/2 ==> 4a₀ = 4k₀L = ln( 3/2 ) ==> k₀ = ln( 3/2 ) / 4L ==> k₀ = ln( 3/2 ) / 2000 ==> k₀L = (1/4)ln( 3/2 )

N(t) = 250 - √(250² - (4 x 10⁴)e^[(1/4)ln(3/2)]t )

N(t) = 250 - √( (2.5² x 10⁴) - (4 x 10⁴)e^[(1/4)ln(3/2)]t )

N(t) = 250 - √( (6.25 x 10⁴) - (4 x 10⁴)e^[(1/4)ln(3/2)]t )

N(t) = 250 - √( (4 x 10⁴)((6.25/4) - e^([(1/4)ln(3/2)]t )

N(t) = 250 - 200√(1.5625 - e^[ln[(3/2)]¹/⁴t )

N(t) = 250 - 200√(1.5625 - e^[ln[(3/2)]ᵀ/⁴ )

N(t) = 250 - 200√(1.5625 - [(3/2)ᵀ/⁴] )

Check: N(0) = 250 - 200√(1.5625 - 1) = 100

Check: N(4) = 250 - 200√(1.5625 - [(3/2)¹/⁴]⁴) = 250 - 200√(1.5625 - 3/2) = 250 - 200(.25) = 250 - 50 = 200