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″ + 7y′ + 12y = 0 with y(0) = 3 and y′(0) = 0. Find y(1).

y″ + 7y′ + 12y = 0y(0) = 3y′(0) = 0

y(1) =

2.Differential equationsHard

A system obeys y″ + 3y′ + 11.25y = 22.5 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find the peak value y_max of its step response.

y″ + 3y′ + 11.25y = 22.5y(0) = 0y′(0) = 0

y_max =

3.IntegralsMedium

Evaluate I = ∫₁^1.5 x·ln x dx.

∫₁^1.5 x ln x dx

I =

4.IntegralsHard

Find the area A of the region enclosed by y = x² and y = 3x.

y = x²y = 3x

A =

5.DerivativesMedium

Let f(x) = x³·e^0.5x. Find f′(2).

f(x) = x³·e^0.5xx = 2

f′(2) =

6.DerivativesHard

The tangent line to f(x) = −x² − 5x + 6 at x = −1 crosses the y-axis at y = b. Find b.

f(x) = −x² − 5x + 6x = −1

b =

7.Numerical methodsMedium

Approximate ∫₀^1.5 sin(x²) dx with the trapezoid rule using n = 3 strips. Give T.

f(x) = sin(x²)[a, b] = [0, 1.5]n = 3

T =

8.Numerical methodsHard

Take one step of the classical Runge–Kutta method (RK4) with h = 0.5 for y′ = e^−ty, y(0) = 0.5. Find y₁ ≈ y(0.5).

y′ = e^−tyy(0) = 0.5h = 0.5

y₁ =

3Solutions

1.Differential equations

Answer: y(1) = 0.4326

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 + 3)(r + 4) = 0 → r = −3, −4

Constants from the initial values

A + B = 3, −3A − 4B = 0

A = 12, B = −9

Evaluate

y(1) = 12·e⁻³ + (−9)·e⁻⁴ = 0.4326

Answer

y(1) = 0.4326

2.Differential equations

Answer: y_max = 2.416

Steady value and roots

y_ss = 2, r = −1.5 ± 3j

Constants from the initial values

y(0) = 2 + A = 0 → A = −2

y′(0) = −1.5A + 3B = 0 → B = −1

Peak

t = π/3 = 1.047

y_max = 2·(1 + e^−1.5·π/3) = 2.416

Answer

y_max = 2.416

3.Integrals

Answer: I = 0.1436

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 = ln x, du = dx/x; dv = x dx, v = x²/2

Antiderivative

∫x ln x dx = (x²/2)ln x − ∫(x/2) dx = (x²/2)ln x − x²/4

Limits

(1.125·ln 1.5 − 0.5625) − (0 − 0.25) = 0.1436

Answer

I = 0.1436

4.Integrals

Answer: A = 4.5

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

x² = 3x → x = 0, x = 3

Integrate upper minus lower

A = [1.5x² − 0.3333x³]₀³ = 4.5

Answer

A = 4.5

5.Derivatives

Answer: f′(2) = 43.49

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^0.5x·(3x² + 0.5·x³)

Substitute

f′(2) = 2.718·(12 + 4) = 43.49

Answer

f′(2) = 43.49

6.Derivatives

Answer: b = 7

Point and slope

f(−1) = 10

f′(x) = −2x − 5 → f′(−1) = −3

Tangent line at x = 0

b = f(x₀) − f′(x₀)·x₀ = 10 − (−3)·(−1) = 7

Answer

b = 7

7.Numerical methods

Answer: T = 0.739

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 + 2·0.247404 + 2·0.841471 + 0.778073 = 2.955823

Trapezoid rule

T = (0.5/2)·2.955823 = 0.739

Answer

T = 0.739

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

8.Numerical methods

Answer: y₁ = 0.9138

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, 0.5) = 1

k₂ = f(0.25, 0.75) = 0.829029

k₃ = f(0.25, 0.707257) = 0.837935

k₄ = f(0.5, 0.918968) = 0.63161

Combine

y₁ = 0.5 + (0.5/6)(1 + 2·0.829029 + 2·0.837935 + 0.63161) = 0.9138

Answer

y₁ = 0.9138

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