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) = 3 and y′(0) = 0. Find y(2).

y″ + 4y′ + 3y = 0y(0) = 3y′(0) = 0

y(2) =

2.Differential equationsHard

A system obeys y″ + 4y′ + 13y = 13 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find y(0.75); angles in radians.

y″ + 4y′ + 13y = 13y(0) = 0y′(0) = 0

y(0.75) =

3.IntegralsMedium

The region under y = √(2x) from x = 0 to x = 3 is rotated about the x-axis. Find the volume V of the solid (a paraboloid).

y = √(2x)0 ≤ x ≤ 3

V =

4.IntegralsHard

Evaluate I = ∫₀^0.5 x²·eˣ dx.

∫₀^0.5 x²eˣ dx

I =

5.DerivativesMedium

Let f(x) = (−2x + 3)⁵. Find f′(1).

f(x) = (−2x + 3)⁵x = 1

f′(1) =

6.DerivativesHard

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

f(x) = 2x² − 2x + 1x = −2

b =

7.Numerical methodsMedium

Approximate ∫₀² √x·e^−x dx with the trapezoid rule using n = 6 strips. Give T.

f(x) = √x·e^−x[a, b] = [0, 2]n = 6

T =

8.Numerical methodsHard

Use Euler's method with h = 0.25 for y′ = cos t − y, y(0) = −1 to estimate y₃ ≈ y(0.75).

y′ = cos t − yy(0) = −1h = 0.25

y₃ =

3Solutions

1.Differential equations

Answer: y(2) = 0.6053

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 = 3, −1A − 3B = 0

A = 4.5, B = −1.5

Evaluate

y(2) = 4.5·e⁻² + (−1.5)·e⁻⁶ = 0.6053

Answer

y(2) = 0.6053

2.Differential equations

Answer: y(0.75) = 1.024

Steady value and roots

y_ss = 1, r = −2 ± 3j

Constants from the initial values

y(0) = 1 + A = 0 → A = −1

y′(0) = −2A + 3B = 0 → B = −0.6667

Evaluate

y(0.75) = 1·(1 − e^−1.5(cos 2.25 + 0.6667·sin 2.25)) = 1.024

Answer

y(0.75) = 1.024

3.Integrals

Answer: V = 28.27

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 2x dx

Integrate

V = π·2·3²/2 = 28.27

Answer

V = 28.27

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) = −10

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

f′(x) = 5·(−2x + 3)⁴·(−2)

Substitute

f′(1) = 5·1⁴·(−2) = −10

Answer

f′(1) = −10

6.Derivatives

Answer: b = −7

Point and slope

f(−2) = 13

f′(x) = 4x − 2 → f′(−2) = −10

Tangent line at x = 0

b = f(x₀) − f′(x₀)·x₀ = 13 − (−10)·(−2) = −7

Answer

b = −7

7.Numerical methods

Answer: T = 0.6149

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.

Nodes and values

h = 0.3333

Σ wᵢfᵢ = 0 + 2·0.41369 + 2·0.419203 + 2·0.367879 + 2·0.304376 + 2·0.243837 + 0.191393 = 3.689364

Trapezoid rule

T = (0.3333/2)·3.689364 = 0.6149

Answer

T = 0.6149

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

8.Numerical methods

Answer: y₃ = 0.1198

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.

Euler steps

y₁ = −1 + 0.25·(2) = −0.5

y₂ = −0.5 + 0.25·(1.468912) = −0.132772

y₃ = −0.132772 + 0.25·(1.010354) = 0.119817

Graph

y₃ = 0.1198

Answer

y₃ = 0.1198

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