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

A forged steel part at 420 °C is left to cool in a workshop at 25 °C. After 28 min its temperature is 122 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). Find the constant k in min⁻¹.

T₀ = 420 °CT_a = 25 °CT(28) = 122 °C

k = min⁻¹

2.Differential equationsHard

A tank holds 400 L of water in which 5 kg of salt is dissolved. Brine with 0.2 kg of salt per litre flows in at 2 L/min, and the well-mixed solution flows out at the same rate. How much salt m is in the tank after 45 min?

V = 400 Lm₀ = 5 kgc = 0.2 kg/Lr = 2 L/min

m = kg

3.IntegralsMedium

The region under y = √(x) from x = 0 to x = 1 is rotated about the x-axis. Find the volume V of the solid (a paraboloid).

y = √(x)0 ≤ x ≤ 1

V =

4.IntegralsHard

Evaluate I = ∫₀^1.5 x²·eˣ dx.

∫₀^1.5 x²eˣ dx

I =

5.DerivativesMedium

Let f(x) = x³·e^−x. Find f′(0.5).

f(x) = x³·e^−xx = 0.5

f′(0.5) =

6.DerivativesHard

Let f(x) = x³ + 9x² + 24x + 5. Find the x-coordinate x of the local maximum.

f(x) = x³ + 9x² + 24x + 5

x =

7.Numerical methodsMedium

The trapezoid rule is used for ∫₀² x³ dx. The error is at most (b − a)³·M/(12n²) with M = max|f″| on [0, 2]. Find the smallest number of strips n that guarantees an error below 0.001.

f(x) = x³[a, b] = [0, 2]ε = 0.001

n =

8.Numerical methodsHard

Use Euler's method with h = 0.5 for y′ = 0.5y, y(0) = 2 to estimate y₄ ≈ y(2).

y′ = 0.5yy(0) = 2h = 0.5

y₄ =

3Solutions

1.Differential equations

Answer: k = 0.05015 min⁻¹

First-order differential equations: dy/dt = k·y has the solution y = y₀·e^kt (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^−28k = 97/395

Solve for k

k = ln(395/97)/28 = 0.05015 min⁻¹

Answer

k = 0.05015 min⁻¹

2.Differential equations

Answer: m = 20.11 kg

Set up the equation

dm/dt = 2·0.2 − (2/400)·m = 0.4 − 0.005·m

Solve

The integrating factor is e^rt/V; the solution approaches the equilibrium c·V.

m(t) = 80 + (−75)·e^−t/200

Substitute t

m(45) = 80 + (−75)·0.7985 = 20.11 kg

Answer

m = 20.11 kg

3.Integrals

Answer: V = 1.571

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∫₀¹ x dx

Integrate

V = π·1·1²/2 = 1.571

Answer

V = 1.571

4.Integrals

Answer: I = 3.602

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

4.482·(1.25) − 1·2 = 3.602

Answer

I = 3.602

5.Derivatives

Answer: f′(0.5) = 0.3791

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·(3x² + (−1)·x³)

Substitute

f′(0.5) = 0.6065·(0.75 + (−0.125)) = 0.3791

Answer

f′(0.5) = 0.3791

6.Derivatives

Answer: x = −4

Derivative

f′(x) = 3x² + 18x + 24 = 3·(x² + 6x + 8)

Solve f′(x) = 0

x = −4 or x = −2

Which one is which

f″(x) = 6x + 18: negative at a maximum, positive at a minimum.

f″(−4) = −6, f″(−2) = 6

x = −4

Answer

x = −4

7.Numerical methods

Answer: n = 90

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 = 6·2 = 12

Solve for n

n² > 8·12/(12·0.001) = 8000

n > 89.44 → n = 90

Answer

n = 90

8.Numerical methods

Answer: y₄ = 4.883

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 = 2 + 0.5·(1) = 2.5

y₂ = 2.5 + 0.5·(1.25) = 3.125

y₃ = 3.125 + 0.5·(1.5625) = 3.90625

y₄ = 3.90625 + 0.5·(1.953125) = 4.882813

Graph

y₄ = 4.883

Answer

y₄ = 4.883

The exact value is y(2) = 5.436564.

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