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
tyy(0) = 3y(0.5) = ?

Solve y″ + 4y′ + 4y = 0 with y(0) = 3 and y′(0) = 1. Find y(0.5).

y″ + 4y′ + 4y = 0y(0) = 3y′(0) = 1
y(0.5) =
2.Differential equationsHard
xy12(1, 0)y(2) = ?

Solve the initial value problem y′ + y/x = 4x for x > 0, with y(1) = 0. Find y(2).

y′ + y/x = 4xy(1) = 0
y(2) =
3.IntegralsMedium
xy3y = √(4x)

The region under y = √(4x) from x = 0 to x = 3 is rotated about the x-axis. Find the volume V of the solid (a paraboloid).

y = √(4x)0 ≤ x ≤ 3
V =
4.IntegralsHard
xyy = 0.5x2​y = 4x

Find the area A of the region enclosed by y = 0.5x2 and y = 4x.

y = 0.5x2y = 4x
A =
5.DerivativesMedium
xyx = 2

Let f(x) = ln(x2 + 4). Find f′(2).

f(x) = ln(x2 + 4)x = 2
f′(2) =
6.DerivativesHard
xy

Let f(x) = x3 + 7.5x2 + 18x + 1. Find the x-coordinate x of the local maximum.

f(x) = x3 + 7.5x2 + 18x + 1
x =
7.Numerical methodsMedium
xy00.51

Approximate ∫01 sin(x²) dx with Simpson's rule using n = 2 strips. Give S.

f(x) = sin(x²)[a, b] = [0, 1]n = 2
S =
8.Numerical methodsHard
00.20.411.52ty(0, 1)

Use Euler's method with h = 0.1 for y′ = t² + y², y(0) = 1 to estimate y5 ≈ y(0.5).

y′ = t² + y²y(0) = 1h = 0.1
y5 =

3Solutions

1.Differential equations

Answer: y(0.5) = 2.391

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.

Double root

(r + 2)² = 0 → r = −2

Constants

A = y(0) = 3
y′(0) = B − 2A → B = 1 + 6 = 7

Evaluate

y(0.5) = (3 + 7·0.5)·e−1 = 2.391

Answer

y(0.5) = 2.391
2.Differential equations

Answer: y(2) = 4.667

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.

Integrating factor

μ = e∫1/x dx = x1

Integrate

(x1y)′ = 4x2 → x1y = 4x3/3 + C

Initial condition

0 = 4/3 + C → C = −1.333

Evaluate

y(2) = (4·23/3 + (−1.333))/21 = 4.667

Answer

y(2) = 4.667
3.Integrals

Answer: V = 56.55

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∫₀3 4x dx

Integrate

V = π·4·3²/2 = 56.55

Answer

V = 56.55
4.Integrals

Answer: A = 42.67

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

0.5x2 = 4x → x = 0, x = 8

Integrate upper minus lower

A = [2x2 − 0.1667x3]08 = 42.67

Answer

A = 42.67
5.Derivatives

Answer: f′(2) = 0.5

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 + 4)

Substitute

f′(2) = 4/8 = 0.5

Answer

f′(2) = 0.5
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 + 15x + 18 = 3·(x2 + 5x + 6)

Solve f′(x) = 0

x = −3 or x = −2

Which one is which

f″(x) = 6x + 15: negative at a maximum, positive at a minimum.

f″(−3) = −3, f″(−2) = 3
x = −3

Answer

x = −3
7.Numerical methods

Answer: S = 0.3052

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)weight000110.50.2474044210.8414711

Nodes and values

h = 0.5
Σ wᵢfᵢ = 0 + 4·0.247404 + 0.841471 = 1.831087

Simpson's rule

S = (0.5/3)·1.831087 = 0.3052

Answer

S = 0.3052

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

8.Numerical methods

Answer: y5 = 1.837

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₊₁00111.110.11.11.221.22220.21.2221.5332841.37532830.31.3753281.9815281.57348140.41.5734812.6358431.837066

Euler steps

y₁ = 1 + 0.1·(1) = 1.1
y₂ = 1.1 + 0.1·(1.22) = 1.222
y₃ = 1.222 + 0.1·(1.533284) = 1.375328
y₄ = 1.375328 + 0.1·(1.981528) = 1.573481
y₅ = 1.573481 + 0.1·(2.635843) = 1.837066
00.20.411.52ty

Graph

y5 = 1.837

Answer

y5 = 1.837

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