Differential equations flashcards

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First-order differential equation
An equation connecting x, y and dy/dx (no higher derivatives).
General solution of a first-order differential equation
A solution containing one arbitrary constant; it describes a whole family of curves.
Particular solution
The member of the family fixed by a given condition (e.g. y = 2 when x = 1), with the constant found.
Separable differential equation
One that can be written dy/dx = f(x)g(y): a function of x times a function of y.
Newton's law of cooling
The rate of change of a body's temperature is proportional to the difference between its temperature and that of its surroundings.
Slope field
Short segments drawn at many points, each with the gradient given by the differential equation; solution curves follow them.
Method of separating the variables
dy/dx = f(x)g(y) ⇒ ∫ 1/g(y) dy = ∫ f(x) dx, + one constant. (No — learn it)
Removing ln from ln|y| = f(x) + c
y = A e^(f(x)), where A = ±e^c. (No — derive it)
Solution of dN/dt = kN
N = N₀e^(kt) (growth if k > 0, decay if k < 0). (No — derive it)
Newton's law of cooling (equation and solution)
dθ/dt = −k(θ − θₛ) ⇒ θ = θₛ + (θ₀ − θₛ)e^(−kt). (No — derive it)
Draining tank dh/dt = −k√h — solution
2√h = −kt + c: √h falls linearly, so the tank empties in finite time. (No — derive it)
∫ 1/(a − x) dx
−ln|a − x| + c (inner derivative −1). (No — learn it)
∫ tan x dx
ln|sec x| + c (Yes — in the formulae booklet)
Shape of θ = θₛ + (θ₀ − θₛ)e^(−kt), θ₀ > θₛ
Starts at θ₀, decreases ever more slowly, horizontal asymptote θ = θₛ.
Shape of P = L ÷ (1 + Be^(−ct))
S-shaped: near-exponential growth at first, fastest at P = L/2, levelling off at the asymptote P = L.
Shape of h = (a − bt)² for a draining tank
Decreasing parabola reaching h = 0 at t = a/b; the rising part beyond is not part of the model.
General solution on a graph
A family of curves, one for each value of the constant; a condition picks the curve through one point.
Why only one constant when separating?
Constants on both sides combine into one (c₂ − c₁); a first-order equation needs exactly one.
Sign of k for a decreasing quantity
Write dm/dt = −km with k > 0 so the rate is negative while m > 0.
Translating “inversely proportional to r”
k ÷ r (e.g. dr/dt = k/r).
Translating “proportional to the difference between θ and 20”
k(θ − 20), with a sign chosen to match whether θ rises or falls.
Finding two constants (c and k) from data
Use t = 0 first (c alone), then the second data point for k.
Long-term behaviour of a model
Let t → ∞; terms like e^(−kt) → 0. State the limit in context (e.g. 'approaches 20 °C').
Why use partial fractions in a differential equation
To integrate a rational function such as 1/(P(L − P)) on either side into logarithms.
Solve dy/dx = 2y, y = 5 when x = 0.
ln|y| = 2x + c → y = Ae^(2x) → y = 5e^(2x).
Solve dy/dx = (2x + 1)e^(−y), y = 0 when x = 0.
∫eʸ dy = ∫(2x + 1) dx → eʸ = x² + x + 1 → y = ln(x² + x + 1).
Solve dy/dx = y sin x, y = 3 when x = 0.
ln y = −cos x + c → y = Ae^(−cos x); A = 3e → y = 3e^(1 − cos x).
m = 40e^(−kt), m = 25 at t = 3. Find k and the half-life.
k = ⅓ ln 1.6; half-life = 3 ln 2 ÷ ln 1.6 = 4.42 (3 s.f.).
θ = 20 + 65e^(−kt), θ = 60 at t = 10. When is θ = 35?
k = (1/10) ln(13/8); 65e^(−kt) = 15 → t = 30.2 (3 s.f.).
dh/dt = −k√h, h = 1.44 at t = 0, h = 1 at t = 10. When is the tank empty?
2√h = 2.4 − 0.04t → empty at t = 60.
P = 6 ÷ (1 + 5e^(−0.4t)). When is P = 4, and what is the long-term value?
e^(−0.4t) = 0.1 → t = 2.5 ln 10 = 5.76 (3 s.f.); P → 6.
ln y = 2x + c ⇒ y = e^(2x) + c?
No: y = e^c·e^(2x) = Ae^(2x). The constant multiplies.
When to add the constant of integration
As soon as the integral signs go — before rearranging or exponentiating.
Newton's law trap: θ = θ₀e^(−kt)?
Wrong — that cools to 0 °C. It is the excess θ − θₛ that decays exponentially.
Rounding k early
Keep k exact (e.g. ⅓ ln 1.6) for later parts; round only final answers.
Interval of validity
State when a model stops applying, e.g. a tank formula only until h = 0.

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