Logistic Differantial Equation

ChaoticLlama

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Dec 11, 2004
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Solve the differential equation \(\displaystyle \L\frac{{dM}}{{dt}} = \frac{3}{5}M(1 - \frac{M}{{200}})\) with initial condition M(0) = 50.

I don't know how to solve this. Clearly you start off with separation of variables, and then anti-differentiate using partial fractions, however the question does not work out for me. Thanks for your help.
 
dM/dt = (3/5)M(1 - M/200)

separate variables ...

dM/[M(1 - M/200)] = (3/5)dt

use partial fractions ...

1/[M(1 - M/200)] = A/M + B/(1 - M/200)

1 = A(1 - M/200) + BM

let M = 0 ... A = 1
let M = 200 ... B = 1/200

1/M + (1/200)/(1 - M/200) dM = (3/5) dt

1/M - (-1/200)/(1 - M/200) dM = (3/5) dt

ln(M) - ln(1 - M/200) = (3/5)t + C

ln[M/(1 - M/200)] = (3/5)t + C

M/(1 - M/200) = e<sup>(3/5)t + C</sup>

M/(1 - M/200) = Ce<sup>(3/5)t</sup>

200M/(200 - M) = Ce<sup>(3/5)t</sup>

M(0) = 50 ...

200(50)/150 = 200/3 = C

200M/(200 - M) = (200/3)e<sup>(3/5)t</sup>

M/(200 - M) = (1/3)e<sup>(3/5)t</sup>

M = (200/3)e<sup>(3/5)t</sup> - (M/3)e<sup>(3/5)t</sup>

M + (M/3)e<sup>(3/5)t</sup> = (200/3)e<sup>(3/5)t</sup>

3M + Me<sup>(3/5)t</sup> = 200e<sup>(3/5)t</sup>

M[3 + e<sup>(3/5)t</sup>] = 200e<sup>(3/5)t</sup>

M = 200e<sup>(3/5)t</sup>/[3 + e<sup>(3/5)t</sup>]

M = 200/[1 + 3e<sup>(-3/5)t</sup>]
 
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