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)13Ta​45 °C67 °C

A cup of coffee at 89 °C is left in a room at 21 °C. After 13 min its temperature is 45 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). How long after the start, t, does it take to reach 67 °C?

T0 = 89 °CTa = 21 °CT(13) = 45 °CT = 67 °C
t = min
2.Differential equationsHard
tyy(0) = 2y(0.25) = ?

Solve y″ + y′ + 9.25y = 0 with y(0) = 2 and y′(0) = 0 (released from rest). Find y(0.25); angles in radians.

y″ + y′ + 9.25y = 0y(0) = 2y′(0) = 0
y(0.25) =
3.IntegralsMedium
xy210

Evaluate I = ∫210 1/x dx.

f(x) = 1/x[2, 10]
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 = 1

Let f(x) = ln(x2 + 2). Find f′(1).

f(x) = ln(x2 + 2)x = 1
f′(1) =
6.DerivativesHard
xy

Let f(x) = x3 − 10.5x2 + 30x + 2. Find the x-coordinate x of the local maximum.

f(x) = x3 − 10.5x2 + 30x + 2
x =
7.Numerical methodsMedium

The trapezoid rule is used for ∫02 sin x dx. The error is at most (b − a)³·M/(12n²) with M = max|f″| on [0, 2]. Find the smallest number of strips n that guarantees an error below 1 × 10⁻⁵.

f(x) = sin x[a, b] = [0, 2]ε = 1 × 10⁻⁵
n =
8.Numerical methodsHard
00.050.10.150.211.11.2ty(0, 1)

Take one step of the classical Runge–Kutta method (RK4) with h = 0.2 for y′ = e−ty, y(0) = 1. Find y₁ ≈ y(0.2).

y′ = e−tyy(0) = 1h = 0.2
y1 =

3Solutions

1.Differential equations

Answer: t = 4.879 min

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.

Cooling constant

k = ln(68/24)/13 = 0.08011 min⁻¹

Time to reach the target

t = ln(68/46)/0.08011 = 4.879 min

Answer

t = 4.879 min
2.Differential equations

Answer: y(0.25) = 1.492

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.

Complex roots

r = −0.5 ± 3j

Constants

A = 2, B = 0.5·2/3 = 0.3333

Evaluate

y(0.25) = e−0.125(2·cos 0.75 + 0.3333·sin 0.75) = 1.492

Answer

y(0.25) = 1.492
3.Integrals

Answer: I = 1.609

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 10 − ln 2) = ln 5 = 1.609

Answer

I = 1.609
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′(1) = 0.6667

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 for the logarithm

f′(x) = 2x/(x2 + 2)

Substitute

f′(1) = 2/3 = 0.6667

Answer

f′(1) = 0.6667
6.Derivatives

Answer: x = 2

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.

Derivative

f′(x) = 3x2 − 21x + 30 = 3·(x2 − 7x + 10)

Solve f′(x) = 0

x = 2 or x = 5

Which one is which

f″(x) = 6x − 21: negative at a maximum, positive at a minimum.

f″(2) = −9, f″(5) = 9
x = 2

Answer

x = 2
7.Numerical methods

Answer: n = 259

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.

Second derivative bound

M = |sin(π/2)| = 1

Solve for n

n² > 8·1/(12·1 × 10⁻⁵) = 66667
n > 258.2 → n = 259

Answer

n = 259
8.Numerical methods

Answer: y1 = 1.179

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, 1) = 1
k₂ = f(0.1, 1.1) = 0.895834
k₃ = f(0.1, 1.089583) = 0.896768
k₄ = f(0.2, 1.179354) = 0.789883

Combine

y₁ = 1 + (0.2/6)(1 + 2·0.895834 + 2·0.896768 + 0.789883) = 1.179

Answer

y1 = 1.179

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