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 4 °C is taken out of the fridge into a room at 23 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.03 min⁻¹, find its temperature T after 29 min.
T0 = 4 °CTa = 23 °Ck = 0.03 min⁻¹t = 29 min
T = °C
2.Differential equationsHard
Solve y′ + 0.5y = 9e−t with y(0) = 3. Find y(0.5).
y′ + 0.5y = 9e−ty(0) = 3
y(0.5) =
3.IntegralsMedium
The region under y = √(x) from x = 0 to x = 2 is rotated about the x-axis. Find the volume V of the solid (a paraboloid).
y = √(x)0 ≤ x ≤ 2
V =
4.IntegralsHard
Find the area A of the region enclosed by y = x2 and y = x.
y = x2y = x
A =
5.DerivativesMedium
Let f(x) = x3·e−0.5x. Find f′(1.5).
f(x) = x3·e−0.5xx = 1.5
f′(1.5) =
6.DerivativesHard
The tangent line to f(x) = −x2 + 2x + 5 at x = 3 crosses the y-axis at y = b. Find b.
f(x) = −x2 + 2x + 5x = 3
b =
7.Numerical methodsMedium
A vehicle's speed is measured every 5 s (table). Estimate the distance s travelled from t = 0 to t = 30 s with Simpson's rule.
h = 5 sn = 6
s = m
8.Numerical methodsHard
Use Heun's method (improved Euler) with h = 0.2 for y′ = y(2 − y), y(0) = 0.5 to estimate y2 ≈ y(0.4).
y′ = y(2 − y)y(0) = 0.5h = 0.2
y2 =
3Solutions
1.Differential equations
Answer: T = 15.04 °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) = 23 + (−19)·e−0.03t
Substitute t
T(29) = 23 + (−19)·0.419 = 15.04 °C
Answer
T = 15.04 °C
2.Differential equations
Answer: y(0.5) = 5.437
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 = e0.5t
(e0.5ty)′ = 9e−0.5t
General solution
y = 9/(−0.5)·e−t + C·e−0.5t = −18·e−t + C·e−0.5t
Initial condition
3 = −18 + C → C = 21
Evaluate
y(0.5) = −18·0.6065 + 21·0.7788 = 5.437
Answer
y(0.5) = 5.437
3.Integrals
Answer: V = 6.283
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∫₀2 x dx
Integrate
V = π·1·2²/2 = 6.283
Answer
V = 6.283
4.Integrals
Answer: A = 0.1667
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 = x → x = 0, x = 1
Integrate upper minus lower
A = [0.5x2 − 0.3333x3]01 = 0.1667
Answer
A = 0.1667
5.Derivatives
Answer: f′(1.5) = 2.391
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′(1.5) = 0.4724·(6.75 + (−1.688)) = 2.391
Answer
f′(1.5) = 2.391
6.Derivatives
Answer: b = 14
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) = 2
f′(x) = −2x + 2 → f′(3) = −4
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = 2 − (−4)·3 = 14
Answer
b = 14
7.Numerical methods
Answer: s = 365 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
8 + 4·6 + 2·12 + 4·11 + 2·14 + 4·18 + 19 = 219
Simpson's rule
s = (5/3)·219 = 365 m
Answer
s = 365 m
8.Numerical methods
Answer: y2 = 0.8488
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.
Heun steps
y₁ = 0.5 + (0.2/2)(0.75 + 0.8775) = 0.66275
y₂ = 0.66275 + (0.2/2)(0.886262 + 0.974401) = 0.848816
Graph
y2 = 0.8488
Answer
y2 = 0.8488
The exact value is y(0.4) = 0.851794.
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