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 340 °C is left to cool in a workshop at 29 °C. After 56 min its temperature is 87 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). Find the constant k in min⁻¹.

T₀ = 340 °CT_a = 29 °CT(56) = 87 °C

k = min⁻¹

2.Differential equationsHard

A tank holds 200 L of water in which 10 kg of salt is dissolved. Brine with 0.25 kg of salt per litre flows in at 10 L/min, and the well-mixed solution flows out at the same rate. How much salt m is in the tank after 30 min?

V = 200 Lm₀ = 10 kgc = 0.25 kg/Lr = 10 L/min

m = kg

3.IntegralsMedium

Evaluate I = ∫₀^π/3 x·cos x dx (x in radians).

∫₀^π/3 x·cos x dx

I =

4.IntegralsHard

Find the area A of the region between the parabola y = 4 − x² and the x-axis.

y = 4 − x²

A =

5.DerivativesMedium

Let f(x) = x³·e^−x. Find f′(1.5).

f(x) = x³·e^−xx = 1.5

f′(1.5) =

6.DerivativesHard

Let f(x) = x³ − 4.5x² + 6x − 5. Find the x-coordinate x of the local minimum.

f(x) = x³ − 4.5x² + 6x − 5

x =

7.Numerical methodsMedium

Approximate ∫₀³ √x·e^−x dx with the trapezoid rule using n = 6 strips. Give T.

f(x) = √x·e^−x[a, b] = [0, 3]n = 6

T =

8.Numerical methodsHard

Use Heun's method (improved Euler) with h = 0.2 for y′ = cos t − y, y(0) = 0.5 to estimate y₃ ≈ y(0.6).

y′ = cos t − yy(0) = 0.5h = 0.2

y₃ =

3Solutions

1.Differential equations

Answer: k = 0.02999 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^−56k = 58/311

Solve for k

k = ln(311/58)/56 = 0.02999 min⁻¹

Answer

k = 0.02999 min⁻¹

2.Differential equations

Answer: m = 41.07 kg

Set up the equation

dm/dt = 10·0.25 − (10/200)·m = 2.5 − 0.05·m

Solve

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

m(t) = 50 + (−40)·e^−t/20

Substitute t

m(30) = 50 + (−40)·0.2231 = 41.07 kg

Answer

m = 41.07 kg

3.Integrals

Answer: I = 0.4069

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 = x, du = dx; dv = cos x dx, v = sin x

Antiderivative

∫x cos x dx = x sin x − ∫sin x dx = x sin x + cos x

Limits

(π/3·sin π/3 + cos π/3) − (0 + 1) = 0.4069

Answer

I = 0.4069

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.

Zeros

4 − x² = 0 → x = ±2

Integrate

A = [4x − x³/3]₋₂² = 5.333 − (−5.333) = 10.67

Answer

A = 10.67

5.Derivatives

Answer: f′(1.5) = 0.7531

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′(1.5) = 0.2231·(6.75 + (−3.375)) = 0.7531

Answer

f′(1.5) = 0.7531

6.Derivatives

Answer: x = 2

Derivative

f′(x) = 3x² − 9x + 6 = 3·(x² − 3x + 2)

Solve f′(x) = 0

x = 1 or x = 2

Which one is which

f″(x) = 6x − 9: negative at a maximum, positive at a minimum.

f″(1) = −3, f″(2) = 3

x = 2

Answer

x = 2

7.Numerical methods

Answer: T = 0.7172

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 = 0.5

Σ wᵢfᵢ = 0 + 2·0.428882 + 2·0.367879 + 2·0.273278 + 2·0.191393 + 2·0.129788 + 0.0862337 = 2.868673

Trapezoid rule

T = (0.5/2)·2.868673 = 0.7172

Answer

T = 0.7172

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

8.Numerical methods

Answer: y₃ = 0.6914

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.

Heun steps

y₁ = 0.5 + (0.2/2)(0.5 + 0.380067) = 0.588007

y₂ = 0.588007 + (0.2/2)(0.39206 + 0.254642) = 0.652677

y₃ = 0.652677 + (0.2/2)(0.268384 + 0.118982) = 0.691413

Graph

y₃ = 0.6914

Answer

y₃ = 0.6914

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