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
A can of soda at 6 °C is taken out of the fridge into a room at 27 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.07 min⁻¹, find its temperature T after 15 min.
T0 = 6 °CTa = 27 °Ck = 0.07 min⁻¹t = 15 min
T = °C
2.Differential equationsHard
A tank holds 100 L of water in which 20 kg of salt is dissolved. Brine with 0.05 kg of salt per litre flows in at 5 L/min, and the well-mixed solution flows out at the same rate. How much salt m is in the tank after 25 min?
V = 100 Lm0 = 20 kgc = 0.05 kg/Lr = 5 L/min
m = kg
3.IntegralsMedium
The region under y = √(4x) from x = 0 to x = 4 is rotated about the x-axis. Find the volume V of the solid (a paraboloid).
y = √(4x)0 ≤ x ≤ 4
V =
4.IntegralsHard
Find the area A of the region enclosed by y = 2x2 and y = 3x.
y = 2x2y = 3x
A =
5.DerivativesMedium
Let f(x) = (3x + 3)4. Find f′(1).
f(x) = (3x + 3)4x = 1
f′(1) =
6.DerivativesHard
The tangent line to f(x) = 0.5x2 + 2x − 6 at x = 1 crosses the y-axis at y = b. Find b.
f(x) = 0.5x2 + 2x − 6x = 1
b =
7.Numerical methodsMedium
Approximate ∫01 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, 1]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) = −1. Find y₁ ≈ y(0.5).
y′ = cos t − yy(0) = −1h = 0.5
y1 =
3Solutions
1.Differential equations
Answer: T = 19.65 °C
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.
Solution of the cooling law
T(t) = 27 + (−21)·e−0.07t
Substitute t
T(15) = 27 + (−21)·0.3499 = 19.65 °C
Answer
T = 19.65 °C
2.Differential equations
Answer: m = 9.298 kg
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.
Set up the equation
dm/dt = 5·0.05 − (5/100)·m = 0.25 − 0.05·m
Solve
The integrating factor is ert/V; the solution approaches the equilibrium c·V.
m(t) = 5 + 15·e−t/20
Substitute t
m(25) = 5 + 15·0.2865 = 9.298 kg
Answer
m = 9.298 kg
3.Integrals
Answer: V = 100.5
Volumes of revolution: rotating the region under y = f(x) from x = a to x = b about the x-axis makes a solid whose cross-section at x is a disc of radius f(x). Adding the discs gives V = π∫ₐᵇ f(x)² dx. When the region lies between an outer curve R(x) and an inner curve r(x), each slice is a washer: V = π∫ₐᵇ (R(x)² − r(x)²) dx. Square the radii before you subtract them.
Disc method
V = π∫₀4 4x dx
Integrate
V = π·4·4²/2 = 100.5
Answer
V = 100.5
4.Integrals
Answer: A = 1.125
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
2x2 = 3x → x = 0, x = 1.5
Integrate upper minus lower
A = [1.5x2 − 0.6667x3]01.5 = 1.125
Answer
A = 1.125
5.Derivatives
Answer: f′(1) = 2592
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) = 4·(3x + 3)3·3
Substitute
f′(1) = 4·63·3 = 2592
Answer
f′(1) = 2592
6.Derivatives
Answer: b = −6.5
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(1) = −3.5
f′(x) = x + 2 → f′(1) = 3
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = −3.5 − 3·1 = −6.5
Answer
b = −6.5
7.Numerical methods
Answer: |E| = 0.009617
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.
Exact value
I = 0.459698
Trapezoid rule
T = (0.5/2)·(0 + 2·0.479426 + 0.841471) = 0.450081
Error
|E| = |0.450081 − 0.459698| = 0.009617
Answer
|E| = 0.009617
Relative error: 2.092%.
8.Numerical methods
Answer: y1 = −0.2318
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.
Four slopes
k₁ = f(0, −1) = 2
k₂ = f(0.25, −0.5) = 1.468912
k₃ = f(0.25, −0.632772) = 1.601684
k₄ = f(0.5, −0.199158) = 1.07674
Combine
y₁ = −1 + (0.5/6)(2 + 2·1.468912 + 2·1.601684 + 1.07674) = −0.2318
Answer
y1 = −0.2318
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