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
t (min)T (°C)27Ta​T = ?

A can of soda at 6 °C is taken out of the fridge into a room at 23 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.03 min⁻¹, find its temperature T after 27 min.

T0 = 6 °CTa = 23 °Ck = 0.03 min⁻¹t = 27 min
T = °C
2.Differential equationsHard
tyK = 1y(1.5) = ?

A system obeys y″ + 3y′ + 11.25y = 11.25 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find y(1.5); angles in radians.

y″ + 3y′ + 11.25y = 11.25y(0) = 0y′(0) = 0
y(1.5) =
3.IntegralsMedium
xy12y = x ln x

Evaluate I = ∫₁2 x·ln x dx.

∫₁2 x ln x dx
I =
4.IntegralsHard
xy1y = 2xy = 2x²

The region between the line y = 2x and the parabola y = 2x² is rotated about the x-axis. Find the volume V of the solid.

y = 2xy = 2x²
V =
5.DerivativesMedium
xyx = 0.5

Let f(x) = x·sin(2x), with x in radians. Find f′(0.5).

f(x) = x·sin(2x)x = 0.5 rad
f′(0.5) =
6.DerivativesHard
xyx = 2

The tangent line to f(x) = 3x2 + 4x − 5 at x = 2 crosses the y-axis at y = b. Find b.

f(x) = 3x2 + 4x − 5x = 2
b =
7.Numerical methodsMedium
xy00.40.81.21.62

Approximate ∫02 √(1 + x³) dx with the trapezoid rule using n = 5 strips. Give T.

f(x) = √(1 + x³)[a, b] = [0, 2]n = 5
T =
8.Numerical methodsHard
00.20.40.60.60.81ty(0, 1)

Solve y′ = t − y, y(0) = 1 with Euler's method and step h = 0.25 up to t = 0.75. The exact solution is y = t − 1 + (y₀ + 1)e−t. Find the global error |E| = |y(0.75) − y3|.

y′ = t − yy(0) = 1h = 0.25
|E| =

3Solutions

1.Differential equations

Answer: T = 15.44 °C

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.

Solution of the cooling law

T(t) = 23 + (−17)·e−0.03t

Substitute t

T(27) = 23 + (−17)·0.4449 = 15.44 °C

Answer

T = 15.44 °C
2.Differential equations

Answer: y(1.5) = 1.074

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.

Steady value and roots

yss = 1, r = −1.5 ± 3j

Constants from the initial values

y(0) = 1 + A = 0 → A = −1
y′(0) = −1.5A + 3B = 0 → B = −0.5

Evaluate

y(1.5) = 1·(1 − e−2.25(cos 4.5 + 0.5·sin 4.5)) = 1.074

Answer

y(1.5) = 1.074
3.Integrals

Answer: I = 0.6363

Integration by parts: ∫u·dv = u·v − ∫v·du, from the product rule. Choose u as the factor that gets simpler when you differentiate it (a power of x, or ln x), and dv as the factor you can integrate (eˣ, sin x, cos x, or x when u = ln x). For a definite integral, evaluate u·v at both limits: ∫ₐᵇ u dv = [u·v]ₐᵇ − ∫ₐᵇ v du. A factor x² needs two rounds.

Choose u and dv

u = ln x, du = dx/x; dv = x dx, v = x²/2

Antiderivative

∫x ln x dx = (x²/2)ln x − ∫(x/2) dx = (x²/2)ln x − x²/4

Limits

(2·ln 2 − 1) − (0 − 0.25) = 0.6363

Answer

I = 0.6363
4.Integrals

Answer: V = 1.676

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

2x = 2x² → x = 0, x = 1

Washer method

V = π∫₀1 ((2x)² − (2x²)²) dx

Integrate

V = π(4·1³/3 − 4·1⁵/5) = π·0.5333 = 1.676

Answer

V = 1.676
5.Derivatives

Answer: f′(0.5) = 1.382

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.

Product and chain rule

f′(x) = sin(2x) + 2x·cos(2x)

Substitute

f′(0.5) = 0.8415 + 1·0.5403 = 1.382

Answer

f′(0.5) = 1.382
6.Derivatives

Answer: b = −17

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(2) = 15
f′(x) = 6x + 4 → f′(2) = 16

Tangent line at x = 0

b = f(x₀) − f′(x₀)·x₀ = 15 − 16·2 = −17

Answer

b = −17
7.Numerical methods

Answer: T = 3.268

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)weight001110.41.031504220.81.229634231.21.651666241.62.25743225231

Nodes and values

h = 0.4
Σ wᵢfᵢ = 1 + 2·1.031504 + 2·1.229634 + 2·1.651666 + 2·2.257432 + 3 = 16.340472

Trapezoid rule

T = (0.4/2)·16.340472 = 3.268

Answer

T = 3.268

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

8.Numerical methods

Answer: |E| = 0.101

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.

ntₙyₙf(tₙ, yₙ)yₙ₊₁001−10.7510.250.75−0.50.62520.50.625−0.1250.59375

Euler steps

y3 = 0.59375

Exact value

y(0.75) = 0.694733

Global error

|E| = |0.694733 − 0.59375| = 0.101
00.20.40.60.60.81ty

Graph

Answer

|E| = 0.101

Halving h roughly halves the error of Euler's method.

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