The Geometric Idea

For a first-order ODE \(x' = f(t, x)\), at every point \((t, x)\) in the plane the ODE prescribes the slope that any solution curve through that point must have.

Nullclines and Isoclines

Nullclines — where \(f(t,x) = 0\), i.e., where the slope field is horizontal.

• Solutions have zero slope on a nullcline and may change from increasing to decreasing across it.
• For an autonomous ODE \(x'=f(x)\): nullclines are horizontal lines \(x=c\) where \(f(c)=0\), which are exactly the equilibrium solutions.

Python: The slope_field Helper

def slope_field(f, t_range, x_range, n=20, ax=None, color='steelblue', alpha=0.7):
    """Draw slope field for x' = f(t,x) using normalized quiver arrows."""
    if ax is None:
        fig, ax = plt.subplots()
    T, X = np.meshgrid(np.linspace(*t_range, n), np.linspace(*x_range, n))
    dT, dX = np.ones_like(T), f(T, X)
    norm = np.sqrt(dT**2 + dX**2); norm[norm == 0] = 1
    ax.quiver(T, X, dT/norm, dX/norm, angles='xy', scale=n*0.8,
              width=0.003, color=color, alpha=alpha, headlength=3, headwidth=3)
    ax.set_xlabel('$t$'); ax.set_ylabel('$x$')
    return ax

Key ideas: - np.meshgrid creates the \((t,x)\) grid - dX = f(T, X) evaluates the slope at every grid point - Arrows are normalized to equal length so large slopes don’t dominate

Example 1 — \(x' = -x + 2t\)

Non-autonomous linear ODE (Logan Example 1.5)

• Nullcline: \(f=0 \Rightarrow x = 2t\) (slope-2 line)
• Above nullcline: \(f < 0\), slopes negative
• Below nullcline: \(f > 0\), slopes positive

Example 2 — Autonomous: \(x' = x(1 - x/4)\)

Equilibria at \(x=0\) (unstable) and \(x=4\) (stable). Slope is constant on horizontal lines.

Example 3 — Nonlinear: \(x' = x(x-t)\)

Two nullclines: \(x=0\) (equilibrium solution) and \(x=t\) (not a solution).

Definition and Recipe

An ODE is separable if it factors as \[x' = f(x)\,g(t).\]

Separation of variables: divide by \(f(x)\), then integrate both sides:

\[\frac{1}{f(x)}\,dx = g(t)\,dt \quad\Longrightarrow\quad \int\frac{1}{f(x)}\,dx = \int g(t)\,dt + C.\]

Example 4 — Exponential Growth/Decay

\[\frac{dx}{dt} = rx, \quad r \text{ constant}\]

Separate: \(\dfrac{1}{x}\,dx = r\,dt\)

Integrate: \(\ln|x| = rt + C \;\Rightarrow\; \boxed{x(t) = Ce^{rt}}\)

With \(x(0)=x_0\): \(x(t) = x_0 e^{rt}\).

t, r, x0_sym = sym.symbols('t r x_0', real=True, positive=True)
x = sym.Function('x')
ode = sym.Eq(x(t).diff(t), r * x(t))
sol_ivp = sym.dsolve(ode, x(t), ics={x(0): x0_sym})
display(Math(sym.latex(sol_ivp)))

Example 5 — The Logistic Equation

\[\frac{dx}{dt} = rx\!\left(1-\frac{x}{K}\right), \quad x(0)=x_0\]

Partial fractions: \(\dfrac{1}{x(K-x)} = \dfrac{1/K}{x}+\dfrac{1/K}{K-x}\)

Integrate: \[\frac{1}{K}\ln\!\left|\frac{x}{K-x}\right| = \frac{r}{K}t + C_1 \;\Rightarrow\; \frac{x}{K-x} = Ce^{rt}\]

Solve and apply IC: \[\boxed{x(t) = \frac{K}{1+\!\left(\dfrac{K}{x_0}-1\right)e^{-rt}}}\]

t_s, r_s, K_s, x0_s = sym.symbols('t r K x_0', positive=True)
x_s = sym.Function('x')
logistic_ode = sym.Eq(x_s(t_s).diff(t_s), r_s*x_s(t_s)*(1 - x_s(t_s)/K_s))
sol_log = sym.dsolve(logistic_ode, x_s(t_s), ics={x_s(0): x0_s})
display(Math(sym.latex(sol_log)))

Example 6 — Non-Elementary Integral

\[\frac{dx}{dt} = \frac{2\sqrt{x}\,e^{-t}}{t}, \quad x(1)=4 \qquad \text{(Logan Ex. 1.18)}\]

Separate and integrate (definite form): \[\int_4^x \frac{dy}{2\sqrt{y}} = \int_1^t \frac{e^{-s}}{s}\,ds \;\Rightarrow\; \sqrt{x}-2 = \int_1^t\frac{e^{-s}}{s}\,ds\]

Right-hand side has no elementary form — but can be plotted numerically: \[x(t) = \left(\int_1^t\frac{e^{-s}}{s}\,ds + 2\right)^{\!2}\]

Singular Solutions

Separating variables requires dividing by \(f(x)\) — only valid when \(f(x) \neq 0\).

Any root \(f(c)=0\) gives a constant solution \(x(t)\equiv c\) that may not appear in the general family for any value of \(C\). Such a solution is called a singular solution.

Normal Form and Integrating Factor

A first-order linear ODE in normal form: \[x' + p(t)\,x = q(t)\]

• \(q(t)\equiv 0\): homogeneous; otherwise inhomogeneous
• \(q(t)\) is called the forcing term or source term

Four-Step Procedure

Structure Theorem

The general solution splits as \(x(t) = x_h(t) + x_p(t)\):

Example 8 — \(x' + \tfrac{1}{t}x = 1\)

\(p(t)=1/t\), \(\;q(t)=1\)

Step 2. \(\mu = e^{\int(1/t)dt} = e^{\ln t} = t\)

Step 3. \((tx)' = t\)

Step 4. \(tx = \tfrac{1}{2}t^2 + C \;\Rightarrow\; x(t) = \dfrac{t}{2} + \dfrac{C}{t}\)

With \(x(1)=3\): \(\;\tfrac{1}{2}+C=3 \;\Rightarrow\; C=\tfrac{5}{2}\), so \(\;x(t)=\dfrac{t}{2}+\dfrac{5}{2t}\).

Example 9 — \(x' + 2x = \sin t\)

\(p(t)=2\), \(\;q(t)=\sin t\)

Integrating factor: \(\mu = e^{2t}\)

\((xe^{2t})' = e^{2t}\sin t\)

Integrate by parts: \[\int e^{2t}\sin t\,dt = e^{2t}\!\left(\tfrac{2}{5}\sin t - \tfrac{1}{5}\cos t\right) + C\]

General solution: \[x(t) = \underbrace{\frac{2}{5}\sin t - \frac{1}{5}\cos t}_{x_p \;\text{(steady state)}} + \underbrace{Ce^{-2t}}_{x_h \;\text{(transient)}}\]

Derivation and Solution

Newton’s law of cooling: rate of temperature change is proportional to the temperature difference with the environment: \[\frac{dT}{dt} = -h(T - T_e), \quad T(0)=T_0, \quad h>0\]

In normal form: \(T' + hT = hT_e\)

Integrating factor: \(\mu = e^{ht}\)

\((Te^{ht})' = hT_e e^{ht}\)

\[\boxed{T(t) = T_e + (T_0 - T_e)\,e^{-ht}}\]

Cooling Curves and Forensics

Variable Ambient Temperature

If the environment is not constant but \(Q(t)\), the ODE becomes: \[T' + hT = hQ(t)\]

General solution via integrating factor: \[T(t) = Ce^{-ht} + e^{-ht}\int_0^t hQ(s)\,e^{hs}\,ds\]

This integral form arises whenever the forcing is time-varying — the same structure appears in circuits and reactors.

Governing ODE

An RC circuit with resistance \(R\), capacitance \(C\), and voltage source \(E(t)\):

\[R\,Q' + \frac{Q}{C} = E(t) \quad\Longrightarrow\quad Q' + \frac{1}{RC}\,Q = \frac{E(t)}{R}\]

Integrating factor \(\mu = e^{t/(RC)}\) gives: \[Q(t) = e^{-t/(RC)}\int_0^t \frac{E(s)}{R}\,e^{s/(RC)}\,ds + Q_0\,e^{-t/(RC)}\]

Piecewise Forcing (Logan Ex. 1.29)

\(R=1\), \(C=1/2\) (\(\tau=1/2\)), \(Q(0)=0\), \(E(t) = 1\) for \(0\leq t<2\), else \(0\).

Piece-by-piece with matching at \(t=2\):

\[Q_1(t) = \tfrac{1}{2}(1-e^{-2t}), \quad 0\leq t<2\] \[Q_2(t) = \tfrac{1}{2}(1-e^{-4})\,e^{-2(t-2)}, \quad t\geq 2\]

Mechanical–Electrical Analogy

The RC and Newton’s cooling equations are mathematically identical:

The Chemostat Model

A well-mixed tank of volume \(V\): chemical at concentration \(C(t)\), inflow \(C_{\text{in}}\) at rate \(q\), outflow at same rate.

Mass balance (rate in \(-\) rate out): \[V\,C' = q\,C_{\text{in}} - q\,C \;\Rightarrow\; C' + \frac{q}{V}\,C = \frac{q}{V}\,C_{\text{in}}\]

With \(C(0)=C_0\): \[\boxed{C(t) = C_{\text{in}} + (C_0 - C_{\text{in}})\,e^{-qt/V}}\]

As \(t\to\infty\): \(C(t)\to C_{\text{in}}\) — stable equilibrium at inflow concentration.

Key Takeaways

• A slope field visualizes \(x'=f(t,x)\) geometrically. Nullclines (\(f=0\)) and isoclines (\(f=k\)) organize the field and identify equilibria.

• An ODE is separable if \(x' = f(x)g(t)\). Write \(\frac{dx}{f(x)} = g(t)\,dt\) and integrate. Watch for singular solutions at roots of \(f\).

• A first-order linear ODE \(x'+p(t)x=q(t)\) is solved by the integrating factor \(\mu=e^{\int p\,dt}\). The solution always decomposes as transient (\(x_h\)) + steady state (\(x_p\)).

• Newton’s cooling, RC circuits, and chemostats are all governed by the same first-order linear ODE — same structure, same solution formula, different physical interpretations.