Stewart Calculus · 8th Edition
Chapter 9: Differential Equations
Every key formula from Stewart Calculus Chapter 9, in one searchable page. Click a card to study it — worked examples included.
9.1–9.2 · MODELING & FIELDS
dy/dt = ky ⇒ exponential growth/decay
Direction field: tiny slopes F(x, y) on a grid — solution curves follow them.
Euler’s method
yn+1 = yn + h · F(xn, yn)
Order & initial conditions
An nth-order DE has n constants in its general solution; y(x₀) = y₀ picks the particular one.
Isoclines
Curve F(x, y) = C: every solution crossing it has slope C — a sketching aid.
WORKED EXAMPLE
Euler for dy/dx = x + y, y(0) = 1, h = 0.5:
y₁ = 1.5, y₂ = 2.5 ⇒ y(1) ≈ 2.5 (exact ≈ 3.44).
y₁ = 1.5, y₂ = 2.5 ⇒ y(1) ≈ 2.5 (exact ≈ 3.44).
9.3 · SEPARABLE EQUATIONS
dy/dx = g(x)h(y) ⇒ ∫ dy/h(y) = ∫ g(x) dx
WORKED EXAMPLE
dy/dx = x/y, y(0) = 2:
y dy = x dx ⇒ y² = x² + 4 ⇒ y = √(x² + 4).
y dy = x dx ⇒ y² = x² + 4 ⇒ y = √(x² + 4).
Lost solutions
Dividing by h(y) can drop the equilibrium solutions h(y) = 0 — check them separately.
WORKED EXAMPLE
dy/dx = −2xy, y(0) = 1:
dy/y = −2x dx ⇒ ln|y| = −x² + C ⇒ y = e−x².
dy/y = −2x dx ⇒ ln|y| = −x² + C ⇒ y = e−x².
9.4 · POPULATION MODELS
Natural growth
dP/dt = kP ⇒ P = P₀ekt
Logistic
dP/dt = kP(1 − P/K)
P(t) = K1 + Ae−kt, A = (K − P₀)/P₀
K = carrying capacity; fastest growth at P = K/2.
WORKED EXAMPLE
dP/dt = 0.1P(1 − P/1000), P(0) = 100:
P = 1000/(1 + 9e−0.1t).
P = 1000/(1 + 9e−0.1t).
Mixing problems
dy/dt = (concin)(flowin) − yV(flowout)
Rate in − rate out, with y = amount in the tank.
WORKED EXAMPLE
Newton’s cooling: dT/dt = k(T − 20), T(0) = 95, T(30) = 70:
T = 20 + 75ekt, e30k = 2/3 ⇒ T(60) ≈ 53.3°C.
T = 20 + 75ekt, e30k = 2/3 ⇒ T(60) ≈ 53.3°C.
9.5 · LINEAR EQUATIONS
y′ + P(x)y = Q(x)
μ = e∫P(x) dx, d/dx[μy] = μQ
WORKED EXAMPLE
y′ + 2y = ex: μ = e2x;
(e2xy)′ = e3x ⇒ y = (1/3)ex + Ce−2x.
(e2xy)′ = e3x ⇒ y = (1/3)ex + Ce−2x.
Bernoulli equations
y′ + P(x)y = Q(x)yn ⇒ u = y1−n linearizes it.
KEY NOTES
- Standard form first: divide by the leading coefficient so y′ has coefficient 1 before computing μ.
WORKED EXAMPLE
100 L tank, 0.5 kg/L brine in at 2 L/min, drains at 2 L/min, y(0) = 0:
dy/dt + y/50 = 1 ⇒ y = 50(1 − e−t/50); y(30) ≈ 22.6 kg.
dy/dt + y/50 = 1 ⇒ y = 50(1 − e−t/50); y(30) ≈ 22.6 kg.
WORKED EXAMPLE
y′ + 2xy = x: μ = ex² ⇒
y ex² = ½ex² + C ⇒ y = ½ + Ce−x².
y ex² = ½ex² + C ⇒ y = ½ + Ce−x².
9.6 · PREDATOR–PREY
dx/dt = ax − bxy (prey)
dy/dt = −cy + dxy (predators)
dy/dt = −cy + dxy (predators)
KEY NOTES
- Populations cycle; phase-plane paths are closed loops.
Equilibrium points
(0, 0) and (c/d, a/b)
Set dx/dt = dy/dt = 0 and solve.
KEY NOTES
- Volterra’s principle: harvesting both species (rate ∝ population) decreases predators but increases prey on average.
WORKED EXAMPLE
dx/dt = 0.08x − 0.001xy, dy/dt = −0.02y + 0.00002xy:
equilibrium at x = 1000, y = 80 prey / predators.
equilibrium at x = 1000, y = 80 prey / predators.
WATCH OUT!
Mistakes that cost points
- Separable: +C right after integrating, then solve for y
- Logistic: K is the CEILING, not P₀
- μ multiplies BOTH sides — including Q
- Euler: use small h; errors accumulate
- A direction field is not the solution — follow the slopes
- Linear DE: get STANDARD FORM (y′ coefficient 1) before finding μ
- Separable: h(y) = 0 solutions survive division — don’t lose them
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