Differential Equations: cram pack

Formula sheets, the key method steps for every topic and 8 practice problems, with full solutions at the back.

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1Formulas and methods

Differential equations

Exponential growth/decay
dy/dt = ky → y = y0ekt
Half-life
k = ln 2 / t½
Newton's law of cooling
T(t) = Ta + (T0 − Ta)e−kt
Integrating factor
y′ + p(x)y = q(x): μ = e∫p dx, (μy)′ = μq
Mixing tank
dm/dt = r·c − (r/V)·m

Integrals

Power rule
∫xⁿ dx = xⁿ⁺¹/(n + 1) + C
Definite integral
∫ab f dx = F(b) − F(a)
Common integrals
∫eˣ dx = eˣ, ∫1/x dx = ln|x|, ∫sin x dx = −cos x
Area between curves
∫ab (f − g) dx

Method

  1. Rewrite the integrand as a sum of simple terms where possible.
  2. Apply the power rule: ∫xⁿ dx = xⁿ⁺¹/(n + 1), and standard integrals for ex, 1/x, sin and cos.
  3. Use substitution when you see a function and its derivative together.
  4. For a definite integral, evaluate F(b) − F(a).

Derivatives

Power rule
(xⁿ)′ = n·xⁿ⁻¹
Product rule
(uv)′ = u′v + uv′
Quotient rule
(u/v)′ = (u′v − uv′)/v²
Chain rule
f(g(x))′ = f′(g(x))·g′(x)
Common derivatives
(eˣ)′ = eˣ, (ln x)′ = 1/x, (sin x)′ = cos x, (cos x)′ = −sin x

Method

  1. Identify the structure: sum, product, quotient or composition.
  2. Apply the power rule to each power of x.
  3. Use the product, quotient or chain rule where needed.
  4. Simplify, and evaluate at the point if asked.

Numerical methods

Bisection
c = (a + b)/2, width after n steps (b − a)/2ⁿ
Newton–Raphson
xn+1 = xₙ − f(xₙ)/f′(xₙ)
Linear interpolation
y = y₁ + (y₂ − y₁)(x − x₁)/(x₂ − x₁)
Trapezoid rule
T = (h/2)[f₀ + 2f₁ + … + 2fn−1 + fₙ], h = (b − a)/n
Simpson’s rule (n even)
S = (h/3)[f₀ + 4f₁ + 2f₂ + … + 4fn−1 + fₙ]
Trapezoid error bound
|E| ≤ (b − a)h²M/12, M = max|f″|
Euler
yn+1 = yₙ + h·f(tₙ, yₙ)
Heun
yn+1 = yₙ + (h/2)(k₁ + k₂), k₂ = f(tₙ + h, yₙ + hk₁)
RK4
yn+1 = yₙ + (h/6)(k₁ + 2k₂ + 2k₃ + k₄)

Method

  1. For a root: check the sign change for bisection, or compute f and f′ for Newton, and keep six or more digits in a table of iterates.
  2. For a table value: interpolate linearly between the two neighbouring points.
  3. For an integral: find h = (b − a)/n, list f at every node, apply the weights (1, 2, …, 2, 1 for trapezoid; 1, 4, 2, …, 4, 1 for Simpson with even n) and multiply by h/2 or h/3.
  4. For an ODE: step from t₀ with yₙ₊₁ = yₙ + h·f(tₙ, yₙ) (Euler) or the averaged slope (Heun), one row per step, until you reach the requested t.

2Practice problems

1.Differential equationsMedium
tyy(0) = 2y(1) = ?

Solve y″ + 6y′ + 9y = 0 with y(0) = 2 and y′(0) = -2. Find y(1).

y″ + 6y′ + 9y = 0y(0) = 2y′(0) = -2
y(1) =
2.Differential equationsHard
t (min)m (kg)30 minm = ?

A tank holds 400 L of water in which 10 kg of salt is dissolved. Brine with 0.2 kg of salt per litre flows in at 20 L/min, and the well-mixed solution flows out at the same rate. How much salt m is in the tank after 30 min?

V = 400 Lm0 = 10 kgc = 0.2 kg/Lr = 20 L/min
m = kg
3.IntegralsMedium
xy24

Evaluate I = ∫24 1/x dx.

f(x) = 1/x[2, 4]
I =
4.IntegralsHard
xy2y = xy = 0.5x²

The region between the line y = x and the parabola y = 0.5x² is rotated about the x-axis. Find the volume V of the solid.

y = xy = 0.5x²
V =
5.DerivativesMedium
xyx = 1

Let f(x) = (−2x − 3)4. Find f′(1).

f(x) = (−2x − 3)4x = 1
f′(1) =
6.DerivativesHard
xyx = 3

The tangent line to f(x) = x2 + 4 at x = 3 crosses the y-axis at y = b. Find b.

f(x) = x2 + 4x = 3
b =
7.Numerical methodsMedium
xy00.16670.33330.50.66670.83331

Approximate ∫01 1/(1 + x³) dx with the trapezoid rule using n = 6 strips. Give T.

f(x) = 1/(1 + x³)[a, b] = [0, 1]n = 6
T =
8.Numerical methodsHard
00.20.400.20.4ty(0, 0)

Take one step of the classical Runge–Kutta method (RK4) with h = 0.5 for y′ = e−ty, y(0) = 0. Find y₁ ≈ y(0.5).

y′ = e−tyy(0) = 0h = 0.5
y1 =

3Solutions

1.Differential equations

Answer: y(1) = 0.2987

Second-order linear equations: for y″ + b·y′ + c·y = 0 try y = ert: this gives the characteristic equation r² + b·r + c = 0. Two real roots r₁ ≠ r₂ give y = A·er₁t + B·er₂t (overdamped); a double root r gives y = (A + Bt)·ert (critically damped); complex roots −α ± jω give y = e−αt(A cos ωt + B sin ωt) (underdamped, oscillating). The initial values y(0) and y′(0) fix A and B. Written as y″ + 2ζωny′ + ωn²y = 0, ωn is the natural frequency and ζ the damping ratio: ζ < 1 oscillates. With a constant input, y″ + by′ + cy = c·K, the solution is the steady value K plus the free response.

Double root

(r + 3)² = 0 → r = −3

Constants

A = y(0) = 2
y′(0) = B − 3A → B = -2 + 6 = 4

Evaluate

y(1) = (2 + 4·1)·e−3 = 0.2987

Answer

y(1) = 0.2987
2.Differential equations

Answer: m = 64.38 kg

First-order differential equations: dy/dt = k·y has the solution y = y₀·ekt (growth for k > 0, decay for k < 0); a half-life T gives k = ln 2/T. Newton's law of cooling, dT/dt = −k·(T − Tₐ), gives T(t) = Tₐ + (T₀ − Tₐ)·e−kt. A linear equation y′ + p(x)·y = q(x) is solved with the integrating factor μ = e∫p dx: (μy)′ = μq, so μy = ∫μq dx + C.

Set up the equation

dm/dt = 20·0.2 − (20/400)·m = 4 − 0.05·m

Solve

The integrating factor is ert/V; the solution approaches the equilibrium c·V.

m(t) = 80 + (−70)·e−t/20

Substitute t

m(30) = 80 + (−70)·0.2231 = 64.38 kg

Answer

m = 64.38 kg
3.Integrals

Answer: I = 0.6931

Definite integrals: ∫ₐᵇ f(x) dx is the signed area between the graph and the x-axis from a to b. Find an antiderivative F (with F′ = f) and evaluate F(b) − F(a). Power rule: ∫xⁿ dx = xⁿ⁺¹/(n + 1). The area between two curves is ∫(upper − lower) dx.

Antiderivative

F(x) = ln x

Evaluate at the limits

I = (ln 4 − ln 2) = ln 2 = 0.6931

Answer

I = 0.6931
4.Integrals

Answer: V = 3.351

Volumes of revolution: rotating the region under y = f(x) from x = a to x = b about the x-axis makes a solid whose cross-section at x is a disc of radius f(x). Adding the discs gives V = π∫ₐᵇ f(x)² dx. When the region lies between an outer curve R(x) and an inner curve r(x), each slice is a washer: V = π∫ₐᵇ (R(x)² − r(x)²) dx. Square the radii before you subtract them.

Intersection points

1x = 0.5x² → x = 0, x = 2

Washer method

V = π∫₀2 ((1x)² − (0.5x²)²) dx

Integrate

V = π(1·2³/3 − 0.25·2⁵/5) = π·1.067 = 3.351

Answer

V = 3.351
5.Derivatives

Answer: f′(1) = 1000

Differentiation: the derivative f′(x) is the slope of the graph at x. Power rule: (xⁿ)′ = n·xⁿ⁻¹. Product rule: (uv)′ = u′v + uv′. Chain rule: (f(g(x)))′ = f′(g(x))·g′(x). Where f′(x) = 0 the graph has a horizontal tangent: a maximum, minimum or saddle point.

Chain rule

f′(x) = 4·(−2x − 3)3·(−2)

Substitute

f′(1) = 4·(−5)3·(−2) = 1000

Answer

f′(1) = 1000
6.Derivatives

Answer: b = −5

Differentiation: the derivative f′(x) is the slope of the graph at x. Power rule: (xⁿ)′ = n·xⁿ⁻¹. Product rule: (uv)′ = u′v + uv′. Chain rule: (f(g(x)))′ = f′(g(x))·g′(x). Where f′(x) = 0 the graph has a horizontal tangent: a maximum, minimum or saddle point.

Point and slope

f(3) = 13
f′(x) = 2x → f′(3) = 6

Tangent line at x = 0

b = f(x₀) − f′(x₀)·x₀ = 13 − 6·3 = −5

Answer

b = −5
7.Numerical methods

Answer: T = 0.8339

Numerical integration: split [a, b] into n strips of width h = (b − a)/n with nodes x₀ … xₙ. Trapezoid rule: T = (h/2)[f₀ + 2f₁ + 2f₂ + … + 2fn−1 + fₙ]. Simpson's rule (n even): S = (h/3)[f₀ + 4f₁ + 2f₂ + 4f₃ + … + 4fn−1 + fₙ]. The trapezoid error shrinks like h² (bound (b − a)h²·M/12 with M = max|f″|), Simpson's like h⁴, so Simpson is usually far more accurate for the same nodes.

ixᵢf(xᵢ)weight001110.1666670.995392220.3333330.964286230.50.888889240.6666670.771429250.8333330.6334312610.51

Nodes and values

h = 0.1667
Σ wᵢfᵢ = 1 + 2·0.995392 + 2·0.964286 + 2·0.888889 + 2·0.771429 + 2·0.633431 + 0.5 = 10.006852

Trapezoid rule

T = (0.1667/2)·10.006852 = 0.8339

Answer

T = 0.8339

This integral has no elementary antiderivative, which is exactly when a numerical rule is needed.

8.Numerical methods

Answer: y1 = 0.4629

Numerical solution of y′ = f(t, y): Euler's method steps along the slope at the start of each step: yn+1 = yₙ + h·f(tₙ, yₙ). Heun's method (improved Euler) averages the slope at the start and at an Euler prediction: k₁ = f(tₙ, yₙ), k₂ = f(tₙ + h, yₙ + h·k₁), yn+1 = yₙ + (h/2)(k₁ + k₂). The classical Runge–Kutta method (RK4) combines four slopes: yn+1 = yₙ + (h/6)(k₁ + 2k₂ + 2k₃ + k₄). The global error of Euler is proportional to h, Heun's to h², RK4's to h⁴.

Four slopes

k₁ = f(0, 0) = 1
k₂ = f(0.25, 0.25) = 0.939413
k₃ = f(0.25, 0.234853) = 0.942977
k₄ = f(0.5, 0.471489) = 0.789983

Combine

y₁ = 0 + (0.5/6)(1 + 2·0.939413 + 2·0.942977 + 0.789983) = 0.4629

Answer

y1 = 0.4629

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