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 = y₀e^kt
Half-life: k = ln 2 / t_½
Newton's law of cooling: T(t) = T_a + (T₀ − T_a)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: ∫_a^b f dx = F(b) − F(a)
Common integrals: ∫eˣ dx = eˣ, ∫1/x dx = ln|x|, ∫sin x dx = −cos x
Area between curves: ∫_a^b (f − g) dx

Method

Rewrite the integrand as a sum of simple terms where possible.
Apply the power rule: ∫xⁿ dx = xⁿ⁺¹/(n + 1), and standard integrals for e^x, 1/x, sin and cos.
Use substitution when you see a function and its derivative together.
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

Identify the structure: sum, product, quotient or composition.
Apply the power rule to each power of x.
Use the product, quotient or chain rule where needed.
Simplify, and evaluate at the point if asked.

Numerical methods

Bisection: c = (a + b)/2, width after n steps (b − a)/2ⁿ
Newton–Raphson: x_n+1 = xₙ − f(xₙ)/f′(xₙ)
Linear interpolation: y = y₁ + (y₂ − y₁)(x − x₁)/(x₂ − x₁)
Trapezoid rule: T = (h/2)[f₀ + 2f₁ + … + 2f_n−1 + fₙ], h = (b − a)/n
Simpson’s rule (n even): S = (h/3)[f₀ + 4f₁ + 2f₂ + … + 4f_n−1 + fₙ]
Trapezoid error bound: |E| ≤ (b − a)h²M/12, M = max|f″|
Euler: y_n+1 = yₙ + h·f(tₙ, yₙ)
Heun: y_n+1 = yₙ + (h/2)(k₁ + k₂), k₂ = f(tₙ + h, yₙ + hk₁)
RK4: y_n+1 = yₙ + (h/6)(k₁ + 2k₂ + 2k₃ + k₄)

Method

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.
For a table value: interpolate linearly between the two neighbouring points.
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.
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

Solve y″ + 4y′ + 3y = 0 with y(0) = 2 and y′(0) = 4. Find y(1).

y″ + 4y′ + 3y = 0y(0) = 2y′(0) = 4

y(1) =

2.Differential equationsHard

Solve y″ + 3y′ + 27.25y = 0 with y(0) = 3 and y′(0) = 0 (released from rest). Find y(0.75); angles in radians.

y″ + 3y′ + 27.25y = 0y(0) = 3y′(0) = 0

y(0.75) =

3.IntegralsMedium

Evaluate I = ∫₀¹ 2x·(x² + 1)² dx.

f(x) = 2x·(x² + 1)²[0, 1]

I =

4.IntegralsHard

The region between the line y = x and the parabola y = 0.5x² is rotated about the x-axis. Find the volume V of the solid.

y = xy = 0.5x²

V =

5.DerivativesMedium

Let f(x) = ln(x² + 4). Find f′(3).

f(x) = ln(x² + 4)x = 3

f′(3) =

6.DerivativesHard

The tangent line to f(x) = 2x² − x + 6 at x = 3 crosses the y-axis at y = b. Find b.

f(x) = 2x² − x + 6x = 3

b =

7.Numerical methodsMedium

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

f(x) = 1/x[a, b] = [1, 2]ε = 1 × 10⁻⁵

n =

8.Numerical methodsHard

Take one step of the classical Runge–Kutta method (RK4) with h = 0.1 for y′ = t² + y², y(0) = 1. Find y₁ ≈ y(0.1).

y′ = t² + y²y(0) = 1h = 0.1

y₁ =

3Solutions

1.Differential equations

Answer: y(1) = 1.69

Second-order linear equations: for y″ + b·y′ + c·y = 0 try y = e^rt: this gives the characteristic equation r² + b·r + c = 0. Two real roots r₁ ≠ r₂ give y = A·e^r₁t + B·e^r₂t (overdamped); a double root r gives y = (A + Bt)·e^rt (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″ + 2ζω_ny′ + ω_n²y = 0, ω_n 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.

Characteristic roots

(r + 1)(r + 3) = 0 → r = −1, −3

Constants from the initial values

A + B = 2, −1A − 3B = 4

A = 5, B = −3

Evaluate

y(1) = 5·e⁻¹ + (−3)·e⁻³ = 1.69

Answer

y(1) = 1.69

2.Differential equations

Answer: y(0.75) = −0.9662

Complex roots

r = −1.5 ± 5j

Constants

A = 3, B = 1.5·3/5 = 0.9

Evaluate

y(0.75) = e^−1.125(3·cos 3.75 + 0.9·sin 3.75) = −0.9662

Answer

y(0.75) = −0.9662

3.Integrals

Answer: I = 2.333

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.

Substitution

u = x² + 1, du = 2x dx

I = ∫₁² u² du

Integrate and evaluate

I = [u³/3]₁² = (8 − 1)/3 = 2.333

Answer

I = 2.333

4.Integrals

Answer: V = 3.351

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

1x = 0.5x² → x = 0, x = 2

Washer method

V = π∫₀² ((1x)² − (0.5x²)²) dx

Integrate

V = π(1·2³/3 − 0.25·2⁵/5) = π·1.067 = 3.351

Answer

V = 3.351

5.Derivatives

Answer: f′(3) = 0.4615

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′(3) = 6/13 = 0.4615

Answer

f′(3) = 0.4615

6.Derivatives

Answer: b = −12

Point and slope

f(3) = 21

f′(x) = 4x − 1 → f′(3) = 11

Tangent line at x = 0

b = f(x₀) − f′(x₀)·x₀ = 21 − 11·3 = −12

Answer

b = −12

7.Numerical methods

Answer: n = 130

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 + … + 2f_n−1 + fₙ]. Simpson's rule (n even): S = (h/3)[f₀ + 4f₁ + 2f₂ + 4f₃ + … + 4f_n−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 = 2/1³ = 2

Solve for n

n² > 1·2/(12·1 × 10⁻⁵) = 16667

n > 129.1 → n = 130

Answer

n = 130

8.Numerical methods

Answer: y₁ = 1.111

Numerical solution of y′ = f(t, y): Euler's method steps along the slope at the start of each step: y_n+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), y_n+1 = yₙ + (h/2)(k₁ + k₂). The classical Runge–Kutta method (RK4) combines four slopes: y_n+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(0.05, 1.05) = 1.105

k₃ = f(0.05, 1.05525) = 1.116053

k₄ = f(0.1, 1.111605) = 1.245666

Combine

y₁ = 1 + (0.1/6)(1 + 2·1.105 + 2·1.116053 + 1.245666) = 1.111

Answer

y₁ = 1.111

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