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)30Ta​135 °C

A forged steel part at 300 °C is left to cool in a workshop at 22 °C. After 30 min its temperature is 135 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). Find the constant k in min⁻¹.

T0 = 300 °CTa = 22 °CT(30) = 135 °C
k = min⁻¹
2.Differential equationsHard
tyy(0) = 10y(1.5) = ?

Solve y″ + 3y′ + 11.25y = 0 with y(0) = 10 and y′(0) = 0 (released from rest). Find y(1.5); angles in radians.

y″ + 3y′ + 11.25y = 0y(0) = 10y′(0) = 0
y(1.5) =
3.IntegralsMedium
xy12.5y = x ln x

Evaluate I = ∫₁2.5 x·ln x dx.

∫₁2.5 x ln x dx
I =
4.IntegralsHard
xyy = x2​y = 4x

Find the area A of the region enclosed by y = x2 and y = 4x.

y = x2y = 4x
A =
5.DerivativesMedium
xyx = 0.5

Let f(x) = x3·e2x. Find f′(0.5).

f(x) = x3·e2xx = 0.5
f′(0.5) =
6.DerivativesHard
xyx = 2

The tangent line to f(x) = 0.5x2 − 5x − 5 at x = 2 crosses the y-axis at y = b. Find b.

f(x) = 0.5x2 − 5x − 5x = 2
b =
7.Numerical methodsMedium
xy00.751.5

Approximate ∫01.5 e−x² dx with Simpson's rule using n = 2 strips. Give S.

f(x) = e−x²[a, b] = [0, 1.5]n = 2
S =
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′ = 1 + sin(ty), y(0) = 0. Find y₁ ≈ y(0.1).

y′ = 1 + sin(ty)y(0) = 0h = 0.1
y1 =

3Solutions

1.Differential equations

Answer: k = 0.03001 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−30k = 113/278

Solve for k

k = ln(278/113)/30 = 0.03001 min⁻¹

Answer

k = 0.03001 min⁻¹
2.Differential equations

Answer: y(1.5) = −0.7373

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 = −1.5 ± 3j

Constants

A = 10, B = 1.5·10/3 = 5

Evaluate

y(1.5) = e−2.25(10·cos 4.5 + 5·sin 4.5) = −0.7373

Answer

y(1.5) = −0.7373
3.Integrals

Answer: I = 1.551

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

(3.125·ln 2.5 − 1.5625) − (0 − 0.25) = 1.551

Answer

I = 1.551
4.Integrals

Answer: A = 10.67

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.

Intersection points

x2 = 4x → x = 0, x = 4

Integrate upper minus lower

A = [2x2 − 0.3333x3]04 = 10.67

Answer

A = 10.67
5.Derivatives

Answer: f′(0.5) = 2.718

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′ = e2x·(3x2 + 2·x3)

Substitute

f′(0.5) = 2.718·(0.75 + 0.25) = 2.718

Answer

f′(0.5) = 2.718
6.Derivatives

Answer: b = −7

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) = −13
f′(x) = x − 5 → f′(2) = −3

Tangent line at x = 0

b = f(x₀) − f′(x₀)·x₀ = −13 − (−3)·2 = −7

Answer

b = −7
7.Numerical methods

Answer: S = 0.8461

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)weight001110.750.569783421.50.1053991

Nodes and values

h = 0.75
Σ wᵢfᵢ = 1 + 4·0.569783 + 0.105399 = 3.384531

Simpson's rule

S = (0.75/3)·3.384531 = 0.8461

Answer

S = 0.8461

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

8.Numerical methods

Answer: y1 = 0.1003

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) = 1.0025
k₃ = f(0.05, 0.050125) = 1.002506
k₄ = f(0.1, 0.100251) = 1.010025

Combine

y₁ = 0 + (0.1/6)(1 + 2·1.0025 + 2·1.002506 + 1.010025) = 0.1003

Answer

y1 = 0.1003

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