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 forged steel part at 320 °C is left to cool in a workshop at 27 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.02 min⁻¹, find its temperature T after 47 min.
T0 = 320 °CTa = 27 °Ck = 0.02 min⁻¹t = 47 min
T = °C
2.Differential equationsHard
A tank holds 500 L of pure water. Brine with 0.05 kg of salt per litre flows in at 2 L/min, and the well-mixed solution flows out at the same rate. How much salt m is in the tank after 40 min?
V = 500 Lm0 = 0 kgc = 0.05 kg/Lr = 2 L/min
m = kg
3.IntegralsMedium
Evaluate I = ∫01 e0.5x dx.
f(x) = e0.5x[0, 1]
I =
4.IntegralsHard
Find the area A of the region enclosed by y = 2x2 and y = 2x.
y = 2x2y = 2x
A =
5.DerivativesMedium
Let f(x) = (3x + 1)3. Find f′(0).
f(x) = (3x + 1)3x = 0
f′(0) =
6.DerivativesHard
Let f(x) = x3 − 9x2 + 24x + 1. Find the x-coordinate x of the local minimum.
f(x) = x3 − 9x2 + 24x + 1
x =
7.Numerical methodsMedium
Approximate ∫01.5 sin(x²) dx with the trapezoid rule using n = 3 strips. Give T.
f(x) = sin(x²)[a, b] = [0, 1.5]n = 3
T =
8.Numerical methodsHard
Solve y′ = t + y, y(0) = 1 with Euler's method and step h = 0.2 up to t = 1. The exact solution is y = −t − 1 + (y₀ + 1)eᵗ. Find the global error |E| = |y(1) − y5|.
y′ = t + yy(0) = 1h = 0.2
|E| =
3Solutions
1.Differential equations
Answer: T = 141.5 °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 + 293·e−0.02t
Substitute t
T(47) = 27 + 293·0.3906 = 141.5 °C
Answer
T = 141.5 °C
2.Differential equations
Answer: m = 3.696 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 = 2·0.05 − (2/500)·m = 0.1 − 0.004·m
Solve
The integrating factor is ert/V; the solution approaches the equilibrium c·V.
m(t) = 25 + (−25)·e−t/250
Substitute t
m(40) = 25 + (−25)·0.8521 = 3.696 kg
Answer
m = 3.696 kg
3.Integrals
Answer: I = 1.297
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) = e0.5x/0.5
Evaluate at the limits
I = (e0.5 − 1)/0.5 = (1.649 − 1)/0.5 = 1.297
Answer
I = 1.297
4.Integrals
Answer: A = 0.3333
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 = 2x → x = 0, x = 1
Integrate upper minus lower
A = [x2 − 0.6667x3]01 = 0.3333
Answer
A = 0.3333
5.Derivatives
Answer: f′(0) = 9
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) = 3·(3x + 1)2·3
Substitute
f′(0) = 3·12·3 = 9
Answer
f′(0) = 9
6.Derivatives
Answer: x = 4
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 − 18x + 24 = 3·(x2 − 6x + 8)
Solve f′(x) = 0
x = 2 or x = 4
Which one is which
f″(x) = 6x − 18: negative at a maximum, positive at a minimum.
f″(2) = −6, f″(4) = 6
x = 4
Answer
x = 4
7.Numerical methods
Answer: T = 0.739
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.
Nodes and values
h = 0.5
Σ wᵢfᵢ = 0 + 2·0.247404 + 2·0.841471 + 0.778073 = 2.955823
Trapezoid rule
T = (0.5/2)·2.955823 = 0.739
Answer
T = 0.739
This integral has no elementary antiderivative, which is exactly when a numerical rule is needed.
8.Numerical methods
Answer: |E| = 0.4599
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
y5 = 2.97664
Exact value
y(1) = 3.436564
Global error
|E| = |3.436564 − 2.97664| = 0.4599
Graph
Answer
|E| = 0.4599
Halving h roughly halves the error of Euler's method.
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