First-Order ODEs Model Growth, Decay, and Balance

A first-order ODE contains an unknown function and its first derivative. These equations are simple enough to solve in important cases yet rich enough to model growth, decay, transport, and systems moving toward equilibrium.

Separable equations

If an equation can be written as

$$\frac{dy}{dt}=g(t)h(y),$$

then, where division by $h(y)$ is valid,

$$\frac{1}{h(y)}\,dy=g(t)\,dt.$$

Integrating both sides produces an implicit or explicit family of solutions.

Linear first-order equations

The standard form is

$$y'+p(t)y=q(t).$$

The integrating factor

$$\mu(t)=e^{\int p(t)\,dt}$$

turns the left side into a product derivative:

$$(\mu y)'=\mu q.$$

Integration then gives

$$y=\frac{1}{\mu}\left(\int \mu q\,dt+C\right).$$

Equilibria and stability

For an autonomous equation $y'=f(y)$, equilibria satisfy $f(y^*)=0$. A phase line shows whether nearby states move toward or away from each equilibrium. Attraction indicates stability; repulsion indicates instability.

Exact and numerical solutions

Not every first-order equation belongs to a solvable symbolic family. Euler’s method uses

$$y_{n+1}=y_n+h f(t_n,y_n)$$

to follow the local slope over small steps. Smaller steps usually improve accuracy but increase work, and stability can impose stricter limits.

Classify before choosing a method

Ask in order: Is the equation autonomous? Separable? Linear after standardization? Does the problem require an explicit formula, or only stability or a numerical trajectory? Method selection should follow structure, not keyword matching.

Check your understanding

Is $y'=t+y$ autonomous, separable, or linear?

Show the reasoning

It is not autonomous because $t$ appears explicitly. It is not separable into $g(t)h(y)$. It is linear: $y'-y=t$.