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″ + 7y′ + 12y = 0 with y(0) = 5 and y′(0) = -4. Find y(1.5).
y″ + 7y′ + 12y = 0y(0) = 5y′(0) = -4
y(1.5) =
2.Differential equationsHard
Solve y′ + 3y = 5e−0.5t with y(0) = 3. Find y(1.5).
y′ + 3y = 5e−0.5ty(0) = 3
y(1.5) =
3.IntegralsMedium
Evaluate I = ∫03 e−0.5x dx.
f(x) = e−0.5x[0, 3]
I =
4.IntegralsHard
Evaluate I = ∫₀1.5 x²·eˣ dx.
∫₀1.5 x²eˣ dx
I =
5.DerivativesMedium
Let f(x) = x·sin(x), with x in radians. Find f′(1).
f(x) = x·sin(x)x = 1 rad
f′(1) =
6.DerivativesHard
Let f(x) = x3 − 3x2 − 9x. Find the x-coordinate x of the local minimum.
f(x) = x3 − 3x2 − 9x
x =
7.Numerical methodsMedium
The trapezoid rule is used for ∫01 eˣ dx. The error is at most (b − a)³·M/(12n²) with M = max|f″| on [0, 1]. Find the smallest number of strips n that guarantees an error below 1 × 10⁻⁵.
f(x) = eˣ[a, b] = [0, 1]ε = 1 × 10⁻⁵
n =
8.Numerical methodsHard
Solve y′ = −2ty, y(0) = 2 with Euler's method and step h = 0.5 up to t = 2. The exact solution is y = y₀e−t². Find the global error |E| = |y(2) − y4|.
y′ = −2tyy(0) = 2h = 0.5
|E| =
3Solutions
1.Differential equations
Answer: y(1.5) = 0.1505
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.
Characteristic roots
(r + 3)(r + 4) = 0 → r = −3, −4
Constants from the initial values
A + B = 5, −3A − 4B = -4
A = 16, B = −11
Evaluate
y(1.5) = 16·e−4.5 + (−11)·e−6 = 0.1505
Answer
y(1.5) = 0.1505
2.Differential equations
Answer: y(1.5) = 0.9558
First-order differential equations: dy/dt = k·y has the solution y = y₀·ekt (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.
Integrating factor
μ = e3t
(e3ty)′ = 5e2.5t
General solution
y = 5/2.5·e−0.5t + C·e−3t = 2·e−0.5t + C·e−3t
Initial condition
3 = 2 + C → C = 1
Evaluate
y(1.5) = 2·0.4724 + 1·0.01111 = 0.9558
Answer
y(1.5) = 0.9558
3.Integrals
Answer: I = 1.554
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) = e−0.5x/(−0.5)
Evaluate at the limits
I = (e−1.5 − 1)/(−0.5) = (0.2231 − 1)/(−0.5) = 1.554
Answer
I = 1.554
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′(1) = 1.382
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 and chain rule
f′(x) = sin(x) + x·cos(x)
Substitute
f′(1) = 0.8415 + 1·0.5403 = 1.382
Answer
f′(1) = 1.382
6.Derivatives
Answer: x = 3
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.
Derivative
f′(x) = 3x2 − 6x − 9 = 3·(x2 − 2x − 3)
Solve f′(x) = 0
x = −1 or x = 3
Which one is which
f″(x) = 6x − 6: negative at a maximum, positive at a minimum.
f″(−1) = −12, f″(3) = 12
x = 3
Answer
x = 3
7.Numerical methods
Answer: n = 151
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.
Second derivative bound
M = e1 = 2.718
Solve for n
n² > 1·2.718/(12·1 × 10⁻⁵) = 22652
n > 150.5 → n = 151
Answer
n = 151
8.Numerical methods
Answer: |E| = 0.03663
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 = 0
Exact value
y(2) = 0.0366313
Global error
|E| = |0.0366313 − 0| = 0.03663
Graph
Answer
|E| = 0.03663
Halving h roughly halves the error of Euler's method.
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