Differential equations
- Exponential growth/decay
- dy/dt = ky → y = y0ekt
- Half-life
- k = ln 2 / t½
- Newton's law of cooling
- T(t) = Ta + (T0 − Ta)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
Differential Equations: cram pack
Formula sheets, the key method steps for every topic and 8 practice problems, with full solutions at the back.
1Formulas and methods
Integrals
- Power rule
- ∫xⁿ dx = xⁿ⁺¹/(n + 1) + C
- Definite integral
- ∫ab f dx = F(b) − F(a)
- Common integrals
- ∫eˣ dx = eˣ, ∫1/x dx = ln|x|, ∫sin x dx = −cos x
- Area between curves
- ∫ab (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 ex, 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
- xn+1 = xₙ − f(xₙ)/f′(xₙ)
- Linear interpolation
- y = y₁ + (y₂ − y₁)(x − x₁)/(x₂ − x₁)
- Trapezoid rule
- T = (h/2)[f₀ + 2f₁ + … + 2fn−1 + fₙ], h = (b − a)/n
- Simpson’s rule (n even)
- S = (h/3)[f₀ + 4f₁ + 2f₂ + … + 4fn−1 + fₙ]
- Trapezoid error bound
- |E| ≤ (b − a)h²M/12, M = max|f″|
- Euler
- yn+1 = yₙ + h·f(tₙ, yₙ)
- Heun
- yn+1 = yₙ + (h/2)(k₁ + k₂), k₂ = f(tₙ + h, yₙ + hk₁)
- RK4
- yn+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″ + 8y′ + 16y = 0 with y(0) = 3 and y′(0) = 4. Find y(0.5).
y″ + 8y′ + 16y = 0y(0) = 3y′(0) = 4
y(0.5) =
2.Differential equationsHard
A system obeys y″ + 2y′ + 5y = 5 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find y(1); angles in radians.
y″ + 2y′ + 5y = 5y(0) = 0y′(0) = 0
y(1) =
3.IntegralsMedium
Evaluate I = ∫24 2/x dx.
f(x) = 2/x[2, 4]
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) = (−2x + 5)5. Find f′(0).
f(x) = (−2x + 5)5x = 0
f′(0) =
6.DerivativesHard
The tangent line to f(x) = 2x2 − 5x + 5 at x = 3 crosses the y-axis at y = b. Find b.
f(x) = 2x2 − 5x + 5x = 3
b =
7.Numerical methodsMedium
A vehicle's speed is measured every 10 s (table). Estimate the distance s travelled from t = 0 to t = 60 s with Simpson's rule.
h = 10 sn = 6
s = m
8.Numerical methodsHard
Solve y′ = t − y, y(0) = 3 with Euler's method and step h = 0.1 up to t = 0.4. The exact solution is y = t − 1 + (y₀ + 1)e−t. Find the global error |E| = |y(0.4) − y4|.
y′ = t − yy(0) = 3h = 0.1
|E| =
3Solutions
1.Differential equations
Answer: y(0.5) = 1.489
Second-order linear equations: for y″ + b·y′ + c·y = 0 try y = ert: this gives the characteristic equation r² + b·r + c = 0. Two real roots r₁ ≠ r₂ give y = A·er₁t + B·er₂t (overdamped); a double root r gives y = (A + Bt)·ert (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″ + 2ζωny′ + ωn²y = 0, ωn 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.
Double root
(r + 4)² = 0 → r = −4
Constants
A = y(0) = 3
y′(0) = B − 4A → B = 4 + 12 = 16
Evaluate
y(0.5) = (3 + 16·0.5)·e−2 = 1.489
Answer
y(0.5) = 1.489
2.Differential equations
Answer: y(1) = 0.9858
Second-order linear equations: for y″ + b·y′ + c·y = 0 try y = ert: this gives the characteristic equation r² + b·r + c = 0. Two real roots r₁ ≠ r₂ give y = A·er₁t + B·er₂t (overdamped); a double root r gives y = (A + Bt)·ert (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″ + 2ζωny′ + ωn²y = 0, ωn 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
yss = 1, r = −1 ± 2j
Constants from the initial values
y(0) = 1 + A = 0 → A = −1
y′(0) = −1A + 2B = 0 → B = −0.5
Evaluate
y(1) = 1·(1 − e−1(cos 2 + 0.5·sin 2)) = 0.9858
Answer
y(1) = 0.9858
3.Integrals
Answer: I = 1.386
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.
Antiderivative
F(x) = 2·ln x
Evaluate at the limits
I = 2·(ln 4 − ln 2) = 2·ln 2 = 1.386
Answer
I = 1.386
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′(0) = −6250
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 + 5)4·(−2)
Substitute
f′(0) = 5·54·(−2) = −6250
Answer
f′(0) = −6250
6.Derivatives
Answer: b = −13
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.
Point and slope
f(3) = 8
f′(x) = 4x − 5 → f′(3) = 7
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = 8 − 7·3 = −13
Answer
b = −13
7.Numerical methods
Answer: s = 576.7 m
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 + … + 2fn−1 + fₙ]. Simpson's rule (n even): S = (h/3)[f₀ + 4f₁ + 2f₂ + 4f₃ + … + 4fn−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.
Weighted sum
9 + 4·8 + 2·10 + 4·8 + 2·7 + 4·12 + 18 = 173
Simpson's rule
s = (10/3)·173 = 576.7 m
Answer
s = 576.7 m
8.Numerical methods
Answer: |E| = 0.05688
Numerical solution of y′ = f(t, y): Euler's method steps along the slope at the start of each step: yn+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), yn+1 = yₙ + (h/2)(k₁ + k₂). The classical Runge–Kutta method (RK4) combines four slopes: yn+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
y4 = 2.0244
Exact value
y(0.4) = 2.08128
Global error
|E| = |2.08128 − 2.0244| = 0.05688
Graph
Answer
|E| = 0.05688
Halving h roughly halves the error of Euler's method.
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