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 320 °C is left to cool in a workshop at 26 °C. After 33 min its temperature is 135 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). Find the constant k in min⁻¹.

T₀ = 320 °CT_a = 26 °CT(33) = 135 °C

k = min⁻¹

2.Differential equationsHard

A system obeys y″ + 4y′ + 20y = 200 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find y(1); angles in radians.

y″ + 4y′ + 20y = 200y(0) = 0y′(0) = 0

y(1) =

3.IntegralsMedium

Evaluate I = ∫₀^π x·sin x dx (x in radians).

∫₀^π x·sin x dx

I =

4.IntegralsHard

Evaluate I = ∫₀³ x²·eˣ dx.

∫₀³ x²eˣ dx

I =

5.DerivativesMedium

Let f(x) = x·e^0.5x. Find f′(1).

f(x) = x·e^0.5xx = 1

f′(1) =

6.DerivativesHard

The tangent line to f(x) = 2x² + x + 5 at x = 1 crosses the y-axis at y = b. Find b.

f(x) = 2x² + x + 5x = 1

b =

7.Numerical methodsMedium

Approximate ∫₀³ sin x dx with the trapezoid rule, n = 2, and compare it with the exact value. Find the absolute error |E| = |T − I|.

f(x) = sin x[a, b] = [0, 3]n = 2

|E| =

8.Numerical methodsHard

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

y′ = cos t − yy(0) = 0.5h = 0.5

y₁ =

3Solutions

1.Differential equations

Answer: k = 0.03007 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^−33k = 109/294

Solve for k

k = ln(294/109)/33 = 0.03007 min⁻¹

Answer

k = 0.03007 min⁻¹

2.Differential equations

Answer: y(1) = 11.4

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.

Steady value and roots

y_ss = 10, r = −2 ± 4j

Constants from the initial values

y(0) = 10 + A = 0 → A = −10

y′(0) = −2A + 4B = 0 → B = −5

Evaluate

y(1) = 10·(1 − e⁻²(cos 4 + 0.5·sin 4)) = 11.4

Answer

y(1) = 11.4

3.Integrals

Answer: I = 3.142

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 = sin x dx, v = −cos x

Antiderivative

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

Limits

(−π·cos π + sin π) − (0 + 0) = 3.142

Answer

I = 3.142

4.Integrals

Answer: I = 98.43

First round

∫x²eˣ dx = x²eˣ − ∫2x eˣ dx

Second round

∫2x eˣ dx = 2x eˣ − 2eˣ

∫x²eˣ dx = eˣ(x² − 2x + 2)

Limits

20.09·(5) − 1·2 = 98.43

Answer

I = 98.43

5.Derivatives

Answer: f′(1) = 2.473

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·(1 + 0.5·x)

Substitute

f′(1) = 1.649·(1 + 0.5) = 2.473

Answer

f′(1) = 2.473

6.Derivatives

Answer: b = 3

Point and slope

f(1) = 8

f′(x) = 4x + 1 → f′(1) = 5

Tangent line at x = 0

b = f(x₀) − f′(x₀)·x₀ = 8 − 5·1 = 3

Answer

b = 3

7.Numerical methods

Answer: |E| = 0.3879

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.

Exact value

I = 1.989992

Trapezoid rule

T = (1.5/2)·(0 + 2·0.997495 + 0.14112) = 1.602082

Error

|E| = |1.602082 − 1.989992| = 0.3879

Answer

|E| = 0.3879

Relative error: 19.49%.

8.Numerical methods

Answer: y₁ = 0.6783

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) = 0.5

k₂ = f(0.25, 0.625) = 0.343912

k₃ = f(0.25, 0.585978) = 0.382934

k₄ = f(0.5, 0.691467) = 0.186115

Combine

y₁ = 0.5 + (0.5/6)(0.5 + 2·0.343912 + 2·0.382934 + 0.186115) = 0.6783

Answer

y₁ = 0.6783

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