Separation of Variables Math Example 2

Follow the full solution, then compare it with the other examples linked below.

Example 2

hard
Solve the logistic DE dy/dt=y(1−y)dy/dt = y(1-y) with y(0)=1/2y(0) = 1/2.

Solution

  1. 1
    Partial fractions: ∫dyy(1−y)=∫(1/y+1/(1−y)) dy=t+C\int\frac{dy}{y(1-y)} = \int(1/y + 1/(1-y))\,dy = t+C.
  2. 2
    ln⁡∣y∣−ln⁡∣1−y∣=t+C\ln|y|-\ln|1-y| = t+C, so y/(1−y)=Kety/(1-y) = Ke^t.
  3. 3
    y(0)=1/2y(0)=1/2: K=1K=1. Solve: y=et1+et=11+e−ty = \frac{e^t}{1+e^t} = \frac{1}{1+e^{-t}}.

Answer

y=11+e−ty = \frac{1}{1+e^{-t}}
Logistic equations use partial fractions in the separation step. The solution is the sigmoid function, modeling bounded growth.

About Separation of Variables

A method for solving first-order DEs of the form dydx=f(x)⋅g(y)\frac{dy}{dx} = f(x) \cdot g(y): rearrange to dyg(y)=f(x) dx\frac{dy}{g(y)} = f(x)\,dx, then integrate both sides.

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More Separation of Variables Examples

Example 1 easy

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Example 4 medium

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