Interpreting ODE Solutions

Interpreting Solutions to Differential Equations

General vs Particular Solutions

Example: $\dfrac{dy}{dx} = 3x^2$ has general solution $y = x^3 + C$.
With IC $y(1) = 5$: \$5 = 1 + C \Rightarrow C = 4$. Particular solution: $y = x^3 + 4$.

Interpreting the Constant $C$

In models, $C$ typically encodes the initial state of the system:

• In $y = Ae^{kt}$: $A = y(0)$ is the initial value.
• In $T(t) = T_a + (T_0 - T_a)e^{-kt}$: $T_0$ is the initial temperature.
• In population models: $C$ reflects starting population or initial perturbation.

Long-Term Behaviour (Equilibrium)

An equilibrium solution is a constant function $y = c$ satisfying the ODE (i.e., $dy/dx = 0$).

Example: $\dfrac{dy}{dx} = 2y(y-3)$. Equilibria: $y = 0$ and $y = 3$.

A solution is stable if nearby solutions converge to it; unstable if they diverge.

For $\dfrac{dy}{dt} = r(K-y)$ (where $r, K > 0$): as $t \to \infty$, $y \to K$ (stable equilibrium).

Reading Direction Fields

A direction field (slope field) plots short line segments with slope $f(x,y)$ at each grid point $(x,y)$. Any particular solution is a curve tangent to these segments at each point.

To sketch a solution:
1. Identify isoclines: curves where $dy/dx = k$ (constant).
2. Trace the curve starting from the IC, following the slope directions.

Validity of Solutions

Always check:
- Domain of definition: where the solution is valid (e.g., $y > 0$ if $\ln y$ appears)
- Physical constraints: concentrations cannot be negative; populations must be non-negative
- Blow-up: does the solution $\to \pm\infty$ at a finite $x$? (e.g., $dy/dx = y^2 \Rightarrow y = 1/(C-x)$)

Example: $\dfrac{dy}{dx} = y^2$, $y(0) = 1$.

General solution: $y = \dfrac{1}{C - x}$. IC: $C = 1$. Particular solution: $y = \dfrac{1}{1-x}$.

Valid for $x < 1$ (blows up as $x \to 1^-$).

Worked Interpretation Example

A tank holds 100 L of brine. Brine with concentration 0.2 kg/L flows in at 5 L/min; well-mixed solution flows out at 5 L/min. The amount of salt $S(t)$ (kg) satisfies:
$$\frac{dS}{dt} = 0.2 \times 5 - \frac{S}{100} \times 5 = 1 - \frac{S}{20}$$

Equilibrium: \$1 - S/20 = 0 \Rightarrow S = 20$ kg (steady-state concentration $= 0.2$ kg/L as expected).

Solving: $S(t) = 20 + (S_0 - 20)e^{-t/20}$.

If $S(0) = 0$: $S(t) = 20(1 - e^{-t/20})$. As $t \to \infty$, $S \to 20$. \checkmark

KEY TAKEAWAY: The general solution is a family of curves; an initial condition pins down one specific curve. Always state what the constant represents in context.

EXAM TIP: When asked to “interpret” a solution, comment on the initial value, long-term behaviour/equilibrium, and any physical constraints or domain restrictions.

COMMON MISTAKE: Finding the general solution and forgetting to substitute the initial condition, leaving $C$ undetermined in a question that specified initial values.