Ordinary Differential Equations Describe Evolving Systems

Many scientific laws do not state a quantity directly. They state how that quantity changes. An ordinary differential equation, or ODE, connects an unknown function of one independent variable to its derivatives.

What makes an equation differential?

The equation

$$\frac{dy}{dt}=ky$$

says that the rate of change of $y$ is proportional to its current value. The unknown is not a number but a function $y(t)$. Solving means finding functions whose derivatives satisfy the equation.

Families and initial conditions

The growth equation has the family of solutions

$$y(t)=Ce^{kt}.$$

The constant $C$ is determined by additional information. If $y(0)=y_0$, then $C=y_0$ and the initial-value problem has the solution $y(t)=y_0e^{kt}$.

An $n$th-order equation generally needs $n$ independent conditions to select one solution, though existence and uniqueness still depend on the equation.

Three ways to understand a solution

An analytic solution is an explicit formula. A qualitative analysis studies direction fields, equilibria, and long-term behavior. A numerical solution approximates values step by step. Most real models need all three viewpoints.

Modeling requires interpretation

Every term has units and a physical role. In $y'=ky$, $k$ must have units of inverse time. The sign of $k$ distinguishes growth from decay. Assumptions such as constant $k$ determine where the model is credible.

Verify a candidate solution

For $y'=2ty$, $y(0)=3$, consider $y=3e^{t^2}$. Differentiate:

$y'=3e^{t^2}(2t)=2ty,$

and check $y(0)=3$. Verification requires both the differential equation and the auxiliary condition.

Existence is not the same as a formula

If $f(t,y)$ is continuous near an initial point, $y'=f(t,y)$ has a local solution. If $f$ is also locally Lipschitz in $y$, that solution is locally unique. These results justify following a slope field even when no elementary formula exists.

Check your understanding

Why does $y'=\sqrt{|y|}$ with $y(0)=0$ deserve a uniqueness check?

Show the reasoning

The right side is continuous but not Lipschitz at zero. The zero solution exists, and solutions can also wait at zero before departing, so the initial condition does not select a unique trajectory.