An IVP specifies the value of the solution (and its derivatives) at a single point.
For a first-order ODE: one condition $y(x_0) = y_0$.
For a second-order ODE: two conditions $y(x_0) = y_0$ and $y’(x_0) = y_1$.
These uniquely determine the arbitrary constants in the general solution.
A BVP specifies conditions at two different points (the boundaries of a domain).
Example: $y’’ + \lambda y = 0$, $y(0) = 0$, $y(L) = 0$.
(This models vibration of a string clamped at both ends.)
BVPs can have no solution, a unique solution, or infinitely many solutions, depending on $\lambda$.
Exponential growth/decay:
$$\frac{dy}{dt} = ky \Rightarrow y = y_0 e^{kt}$$
$k > 0$: growth; $k < 0$: decay.
Logistic growth:
$$\frac{dP}{dt} = rP\left(1 - \frac{P}{K}\right) \Rightarrow P(t) = \frac{K}{1+Ae^{-rt}}$$
Approaches carrying capacity $K$ as $t \to \infty$.
Newton’s Law of Cooling:
$$\frac{dT}{dt} = -k(T-T_a) \Rightarrow T(t) = T_a + (T_0 - T_a)e^{-kt}$$
Simple harmonic motion (undamped):
$$\frac{d^2x}{dt^2} = -\omega^2 x \Rightarrow x = A\cos(\omega t) + B\sin(\omega t)$$
Damped oscillation:
$$\frac{d^2x}{dt^2} + 2\gamma\frac{dx}{dt} + \omega_0^2 x = 0$$
Under-damped ($\gamma < \omega_0$): $x = e^{-\gamma t}(A\cos\omega_d t + B\sin\omega_d t)$ where $\omega_d = \sqrt{\omega_0^2 - \gamma^2}$.
A drug is introduced into the bloodstream at rate $r$ (mg/h) and eliminated at rate proportional to the current amount $Q(t)$ with constant $k$.
$$\frac{dQ}{dt} = r - kQ$$
This is a linear first-order ODE. Integrating factor $\mu = e^{kt}$:
$$\frac{d}{dt}[e^{kt}Q] = re^{kt} \Rightarrow e^{kt}Q = \frac{r}{k}e^{kt} + C \Rightarrow Q = \frac{r}{k} + Ce^{-kt}$$
If $Q(0) = 0$: $C = -r/k$, so $Q(t) = \dfrac{r}{k}(1-e^{-kt})$.
As $t \to \infty$: $Q \to r/k$ (steady-state drug level). Interpretation: the drug reaches an equilibrium where intake equals elimination.
| Context | Rate equation | Solution |
|---|---|---|
| Proportional rate | $dy/dt = ky$ | $y = y_0 e^{kt}$ |
| Constant input, proportional decay | $dy/dt = a-by$ | $y = a/b + Ce^{-bt}$ |
| Oscillatory | $d^2x/dt^2 = -\omega^2 x$ | $A\cos\omega t + B\sin\omega t$ |
| Competing forces | $ma = F(t,v,x)$ | Solve for $v$, then $x$ |
KEY TAKEAWAY: A differential equation model translates a verbal rate description into mathematics. Solve for the general solution, then apply initial/boundary conditions to pin down the specific scenario.
EXAM TIP: When formulating a model from a word problem, identify: what quantity is changing ($y$), what drives the change, and write $dy/dt = $ (rate in) $-$ (rate out).
APPLICATION: Differential equation models arise throughout physics, biology, chemistry, economics, and engineering. The same mathematical structure can describe populations, drug concentrations, electrical circuits, and mechanical systems.
