Separable Differential Equations

A law for change, not a finished function

A differential equation relates an unknown function to one or more of its derivatives. Instead of announcing a formula for a population, temperature, or concentration, it states how that quantity changes under specified conditions. The question is then to find functions that obey the law. This reversal matters: an antiderivative in a differential-equation problem is not simply an algebraic exercise, but a candidate model that must still satisfy the original rate relationship and any given starting condition.

An equation is separable when its variables can be arranged so all expressions involving the dependent variable are paired with $dy$, and all expressions involving the independent variable are paired with $dx$. A typical form is

$$\frac{dy}{dx}=g(x)h(y).$$

Provided the needed quantities are nonzero where we divide, we can write

$$\frac{1}{h(y)}\,dy=g(x)\,dx.$$

This notation is a compact way to organize the chain rule. It does not mean $dy/dx$ has been treated as an ordinary fraction without justification. After integrating, differentiating the answer provides the check that the separated relationship really reproduces the original equation.

Separate, integrate, and keep the constant visible

Consider the growth law

$$\frac{dy}{dt}=0.4y,$$

where $t$ is measured in days and $y$ is a population measured in cells. The number $0.4$ has units $1/\mathrm{day}$, so the right side has units cells per day, matching $dy/dt$. Separating gives

$$\frac{1}{y}\,dy=0.4\,dt.$$

Integrating both sides gives $\ln|y|=0.4t+C$. Exponentiating produces $|y|=e^C e^{0.4t}$. Rather than carrying the absolute-value sign awkwardly, absorb both the sign and the positive factor into a new arbitrary constant $A$:

$$y=Ae^{0.4t}.$$

The constant is not a decorative extra. It represents the whole family of possible initial populations. If the condition is $y(0)=150$ cells, substitution gives $150=Ae^0$, so $A=150$ and $y=150e^{0.4t}$. The starting condition is applied after integration because it selects one member of the family rather than changing the governing law itself.

The zero solution needs separate attention. Dividing by $y$ excluded $y=0$, yet $y(t)=0$ also satisfies $y'=0.4y$. In a population model it may be physically relevant; in another context it may fall outside the intended domain. Whenever separation divides by an expression containing the dependent variable, check whether the excluded value gives an additional constant solution. This small habit prevents losing a valid solution before the calculation begins.

Definite bounds can carry the initial condition

An alternative avoids the unnamed constant by integrating from the initial state directly. If $y(t_0)=y_0$, then a separated equation can be written as

$$\int_{y_0}^{y(t)}\frac{1}{h(u)}\,du =\int_{t_0}^{t}g(s)\,ds.$$

The letters $u$ and $s$ are dummy variables. They prevent the upper bound $y(t)$ from being confused with the integration variable, and they make clear which endpoint changes. For exponential growth from $y(0)=150$, definite integration gives

$$\int_{150}^{y(t)}\frac{1}{u}\,du =\int_0^t0.4\,ds.$$

This yields $\ln\!\left(\frac{y(t)}{150}\right)=0.4t$, followed by the same solution $y(t)=150e^{0.4t}$. Both styles are valid. The indefinite style emphasizes the general solution family, while the definite style makes the initial state part of the accumulated-change statement from the outset.

Interpretation and verification

The expression $y=150e^{0.4t}$ is not automatically credible merely because the algebra is tidy. Differentiate it: $y'=60e^{0.4t}$. Multiplying the original function by $0.4$ gives $0.4y=60e^{0.4t}$, so the rate law is satisfied. Substituting $t=0$ gives $y(0)=150$, so the initial condition is satisfied as well. These two checks test the two different claims made by the model.

The factor $0.4$ tells a relative growth rate, not an added $0.4$ cells per day. As the population increases, the amount added per day also increases. That is why the graph curves upward and why a linear model would eventually disagree. Units make the contrast explicit: a constant addition rate would have units cells per day, whereas $0.4$ here has units per day and must be multiplied by the current population.

For retrieval, solve the initial-value problem $dy/dt=-0.25y$ with $y(0)=80\,\mathrm{mg}$. First predict whether the curve rises or falls and whether it can cross zero. Then separate variables, apply the condition, and differentiate your result to verify the equation. State the units of $0.25$ and explain what the negative sign means in this model. The next lesson will use qualitative reasoning and numerical steps when a closed-form separated solution is unavailable or less useful.