6.2 · Separation of variables, growth, and logistic models

Separate variable factors and integrate

For $\displaystyle \frac{dy}{dx}=g(x)h(y)$, put $\displaystyle dy/h(y)$ on one side and $\displaystyle g(x)dx$ on the other, then integrate. Check solutions with $\displaystyle h(y)=0$ before dividing; these equilibrium solutions may otherwise be lost.

For $\displaystyle y'=2xy$, the nonzero solutions satisfy $\displaystyle dy/y=2x\,dx$. Integration gives $\displaystyle \ln|y|=x^2+C$, or $\displaystyle y=Ce^{x^2}$. The family also includes $\displaystyle y=0$ when the multiplicative constant is zero.

Exponential growth is proportional to amount

The model $\displaystyle P'=kP$ has solution $\displaystyle P(t)=P_0e^{kt}$. Positive $\displaystyle k$ gives growth and negative $\displaystyle k$ gives decay. The constant $\displaystyle k$ has inverse-time units, and the initial amount fixes the coefficient.

If an amount doubles in time $\displaystyle T$, then $\displaystyle k=\ln2/T$. For a half-life $\displaystyle T$, $\displaystyle k=-\ln2/T$. Use logarithms to solve for time; a constant percentage rate is not the same as constant additive growth.

$$P(t)=P_0e^{kt}$$

Logistic growth has a limiting capacity

The BC logistic model $\displaystyle P'=kP(1-P/K)$ has equilibria $\displaystyle 0$ and $\displaystyle K$, with positive constants $\displaystyle k$ and $\displaystyle K$. For $\displaystyle 0<P<K$, the amount grows toward carrying capacity $\displaystyle K$. Above $\displaystyle K$, the rate is negative.

The growth rate is largest at $\displaystyle P=K/2$, because the quadratic rate as a function of $\displaystyle P$ has its maximum there. The logistic solution is $\displaystyle P(t)=K/(1+Ae^{-kt})$, with $\displaystyle A$ determined by the initial condition.