Integral Calculus and Differential Equations › Separation of Variables and Financial Models

For $y'=h(t)g(y)$, separation writes $dy/g(y)=h(t)\,dt$ on intervals where $g(y)\ne0$. Integrating that branch does not recover excluded equilibrium solutions, which must be checked in the original equation.

The deterministic component of the geometric Brownian motion model for stock prices is:

$$\frac{dS}{dt} = \mu S$$

where:

• $S$: stock price at time $t$
• $\mu$: drift rate (expected return per unit time)
• $\frac{dS}{dt}$: rate of change of stock price

For the positive-price branch $S_0>0$, the proportionality to $S$ means percentage changes are constant: a $100 stock and a $200 stock both grow at the same percentage rate $\mu$. Percentage returns make changes comparable across scales, while dollar profit and loss still matter for position sizing, liabilities, and risk limits. A $10 gain on a $100 stock is a 10% return, whereas the same dollar gain on a $1,000 stock is a 1% return. Separating variables:

$$\frac{dS}{S} = \mu \, dt$$

Integrating:

$$\ln S = \mu t + C$$ $$S(t) = S_0 e^{\mu t}$$

Under the geometric Brownian motion convention $dS_t=\mu S_tdt+\sigma S_tdW_t$, this deterministic solution equals the conditional mean $\mathbb{E}[S_t\mid S_0]=S_0e^{\mu t}$. It is not a path that individual prices oscillate or mean-revert around. Log-price variance is $\sigma^2t$, and variance relative to the squared mean is $e^{\sigma^2t}-1$, so these measures increase for nonzero $\sigma$. Absolute price variance need not increase for negative $\mu$. Deviations do not experience a restoring force.

Many vehicles and pieces of equipment lose economic value with age and use, although some physical assets can appreciate. A simple depreciation model assumes value declines in proportion to its current level:

$$\frac{dV}{dt} = -\delta V$$

where:

• $V$: asset value at time $t$
• $\delta$: constant continuous decay rate, in inverse years; its finite one-year loss fraction is $1-e^{-\delta}$
• $-\delta V$: negative because value decreases over time

Assets sharing a proportional decay rate lose the same percentage over a fixed interval, while a larger starting value gives a larger dollar loss. A discrete 15% annual loss removes $4,500 from $30,000 or $2,250 from $15,000. In the continuous model that annual loss corresponds to $\delta=-\ln(0.85)\approx0.16252$ per year, not $\delta=0.15$. The latter gives a one-year loss of about 13.93%. The continuous ODE has solution:

$$V(t) = V_0 e^{-\delta t}$$

where $V_0$ is the initial asset value at $t = 0$.

For $V_0>0$ and $\delta>0$, the asset value decays exponentially, with half-life $t_{1/2} = \ln(2)/\delta$. The half-life formula follows directly from solving $V_0 e^{-\delta t_{1/2}} = V_0/2$, which gives $e^{-\delta t_{1/2}} = 1/2$ and thus $t_{1/2} = \ln(2)/\delta$. Zero value is a separate equilibrium; zero decay has no finite halving time. The following rates are illustrative continuous parameters, not estimated asset-class depreciation rates.