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)10Ta​62 °C

A cup of coffee at 89 °C is left in a room at 20 °C. After 10 min its temperature is 62 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). Find the constant k in min⁻¹.

T0 = 89 °CTa = 20 °CT(10) = 62 °C
k = min⁻¹
2.Differential equationsHard
tyK = 10y(0.5) = ?

A system obeys y″ + 3y′ + 27.25y = 272.5 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find y(0.5); angles in radians.

y″ + 3y′ + 27.25y = 272.5y(0) = 0y′(0) = 0
y(0.5) =
3.IntegralsMedium
xy2y = 0.5x

The region under y = 0.5x from x = 0 to x = 2 is rotated about the x-axis. Find the volume V of the solid (a cone).

y = 0.5x0 ≤ x ≤ 2
V =
4.IntegralsHard
xy0.5y = x²eˣ

Evaluate I = ∫₀0.5 x²·eˣ dx.

∫₀0.5 x²eˣ dx
I =
5.DerivativesMedium
xyx = 1.5

Let f(x) = x·e−x. Find f′(1.5).

f(x) = x·e−xx = 1.5
f′(1.5) =
6.DerivativesHard
xyx = 1

The tangent line to f(x) = −x2 − 2x + 5 at x = 1 crosses the y-axis at y = b. Find b.

f(x) = −x2 − 2x + 5x = 1
b =
7.Numerical methodsMedium
xy00.33330.66671

Approximate ∫01 √x·e−x dx with the trapezoid rule using n = 3 strips. Give T.

f(x) = √x·e−x[a, b] = [0, 1]n = 3
T =
8.Numerical methodsHard
00.51−3−2−1ty(0, −1)

Use Euler's method with h = 0.2 for y′ = y − t², y(0) = −1 to estimate y5 ≈ y(1).

y′ = y − t²y(0) = −1h = 0.2
y5 =

3Solutions

1.Differential equations

Answer: k = 0.04964 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−10k = 42/69

Solve for k

k = ln(69/42)/10 = 0.04964 min⁻¹

Answer

k = 0.04964 min⁻¹
2.Differential equations

Answer: y(0.5) = 12.94

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 = −1.5 ± 5j

Constants from the initial values

y(0) = 10 + A = 0 → A = −10
y′(0) = −1.5A + 5B = 0 → B = −3

Evaluate

y(0.5) = 10·(1 − e−0.75(cos 2.5 + 0.3·sin 2.5)) = 12.94

Answer

y(0.5) = 12.94
3.Integrals

Answer: V = 2.094

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.

Disc method

V = π∫₀2 0.25x² dx

Integrate

V = π·0.25·2³/3 = 2.094

Answer

V = 2.094

Check: a cone of radius 1 and height 2 has (1/3)πr²h = 2.094.

4.Integrals

Answer: I = 0.0609

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.

First round

∫x²eˣ dx = x²eˣ − ∫2x eˣ dx

Second round

∫2x eˣ dx = 2x eˣ − 2eˣ
∫x²eˣ dx = eˣ(x² − 2x + 2)

Limits

1.649·(1.25) − 1·2 = 0.0609

Answer

I = 0.0609
5.Derivatives

Answer: f′(1.5) = −0.1116

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−x·(1 + (−1)·x)

Substitute

f′(1.5) = 0.2231·(1 + (−1.5)) = −0.1116

Answer

f′(1.5) = −0.1116
6.Derivatives

Answer: b = 6

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(1) = 2
f′(x) = −2x − 2 → f′(1) = −4

Tangent line at x = 0

b = f(x₀) − f′(x₀)·x₀ = 2 − (−4)·1 = 6

Answer

b = 6
7.Numerical methods

Answer: T = 0.3389

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)weight000110.3333330.41369220.6666670.4192032310.3678791

Nodes and values

h = 0.3333
Σ wᵢfᵢ = 0 + 2·0.41369 + 2·0.419203 + 0.367879 = 2.033665

Trapezoid rule

T = (0.3333/2)·2.033665 = 0.3389

Answer

T = 0.3389

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

8.Numerical methods

Answer: y5 = −2.763

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₊₁00−1−1−1.210.2−1.2−1.24−1.44820.4−1.448−1.608−1.769630.6−1.7696−2.1296−2.1955240.8−2.19552−2.83552−2.762624

Euler steps

y₁ = −1 + 0.2·(−1) = −1.2
y₂ = −1.2 + 0.2·(−1.24) = −1.448
y₃ = −1.448 + 0.2·(−1.608) = −1.7696
y₄ = −1.7696 + 0.2·(−2.1296) = −2.19552
y₅ = −2.19552 + 0.2·(−2.83552) = −2.762624
00.51−3−2−1ty

Graph

y5 = −2.763

Answer

y5 = −2.763

The exact value is y(1) = −3.154845.

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