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)23Ta​14 °C

A can of soda at 3 °C is taken out of the fridge into a room at 25 °C. After 23 min its temperature is 14 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). Find the constant k in min⁻¹.

T0 = 3 °CTa = 25 °CT(23) = 14 °C
k = min⁻¹
2.Differential equationsHard
tyK = 10y(0.25) = ?

A system obeys y″ + 2y′ + 10y = 100 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find y(0.25); angles in radians.

y″ + 2y′ + 10y = 100y(0) = 0y′(0) = 0
y(0.25) =
3.IntegralsMedium
xy12

Evaluate I = ∫12 1/x dx.

f(x) = 1/x[1, 2]
I =
4.IntegralsHard
xy4y = 4xy = x²

The region between the line y = 4x and the parabola y = x² is rotated about the x-axis. Find the volume V of the solid.

y = 4xy = x²
V =
5.DerivativesMedium
xyx = 0.5

Let f(x) = x3·e−0.5x. Find f′(0.5).

f(x) = x3·e−0.5xx = 0.5
f′(0.5) =
6.DerivativesHard
xy

Let f(x) = x3 − 10.5x2 + 36x. Find the x-coordinate x of the local maximum.

f(x) = x3 − 10.5x2 + 36x
x =
7.Numerical methodsMedium
xy01.53

Approximate ∫03 sin x dx with Simpson's rule, n = 2, and compare it with the exact value. Find the absolute error |E| = |S − I|.

f(x) = sin x[a, b] = [0, 3]n = 2
|E| =
8.Numerical methodsHard
00.050.100.050.1ty(0, 0)

Take one step of the classical Runge–Kutta method (RK4) with h = 0.1 for y′ = e−ty, y(0) = 0. Find y₁ ≈ y(0.1).

y′ = e−tyy(0) = 0h = 0.1
y1 =

3Solutions

1.Differential equations

Answer: k = 0.03014 min⁻¹

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.

Ratio of temperature differences

e−23k = -11/-22

Solve for k

k = ln(-22/-11)/23 = 0.03014 min⁻¹

Answer

k = 0.03014 min⁻¹
2.Differential equations

Answer: y(0.25) = 2.532

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 = 10, r = −1 ± 3j

Constants from the initial values

y(0) = 10 + A = 0 → A = −10
y′(0) = −1A + 3B = 0 → B = −3.333

Evaluate

y(0.25) = 10·(1 − e−0.25(cos 0.75 + 0.3333·sin 0.75)) = 2.532

Answer

y(0.25) = 2.532
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 2 − ln 1) = ln 2 = 0.6931

Answer

I = 0.6931
4.Integrals

Answer: V = 428.9

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

4x = 1x² → x = 0, x = 4

Washer method

V = π∫₀4 ((4x)² − (1x²)²) dx

Integrate

V = π(16·4³/3 − 1·4⁵/5) = π·136.5 = 428.9

Answer

V = 428.9
5.Derivatives

Answer: f′(0.5) = 0.5354

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 rule

f′(x) = u′v + uv′ = e−0.5x·(3x2 + (−0.5)·x3)

Substitute

f′(0.5) = 0.7788·(0.75 + (−0.0625)) = 0.5354

Answer

f′(0.5) = 0.5354
6.Derivatives

Answer: x = 3

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 + 36 = 3·(x2 − 7x + 12)

Solve f′(x) = 0

x = 3 or x = 4

Which one is which

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

f″(3) = −3, f″(4) = 3
x = 3

Answer

x = 3
7.Numerical methods

Answer: |E| = 0.07556

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.

Exact value

I = 1.989992

Simpson's rule

S = (1.5/3)·(0 + 4·0.997495 + 0.14112) = 2.06555

Error

|E| = |2.06555 − 1.989992| = 0.07556

Answer

|E| = 0.07556

Relative error: 3.797%.

8.Numerical methods

Answer: y1 = 0.09967

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(0.05, 0.05) = 0.997503
k₃ = f(0.05, 0.0498752) = 0.997509
k₄ = f(0.1, 0.0997509) = 0.990074

Combine

y₁ = 0 + (0.1/6)(1 + 2·0.997503 + 2·0.997509 + 0.990074) = 0.09967

Answer

y1 = 0.09967

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