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 340 °C is left to cool in a workshop at 30 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.04 min⁻¹, find its temperature T after 17 min.
T0 = 340 °CTa = 30 °Ck = 0.04 min⁻¹t = 17 min
T = °C
2.Differential equationsHard
A tank holds 1000 L of water in which 10 kg of salt is dissolved. Brine with 0.1 kg of salt per litre flows in at 4 L/min, and the well-mixed solution flows out at the same rate. After how many minutes t does the tank contain 55 kg of salt?
V = 1000 Lm0 = 10 kgc = 0.1 kg/Lr = 4 L/min
t = min
3.IntegralsMedium
Evaluate I = ∫26 1/x dx.
f(x) = 1/x[2, 6]
I =
4.IntegralsHard
Evaluate I = ∫₀1 x²·eˣ dx.
∫₀1 x²eˣ dx
I =
5.DerivativesMedium
Let f(x) = ln(x2 + 1). Find f′(2).
f(x) = ln(x2 + 1)x = 2
f′(2) =
6.DerivativesHard
The tangent line to f(x) = x2 − 3x − 1 at x = 2 crosses the y-axis at y = b. Find b.
f(x) = x2 − 3x − 1x = 2
b =
7.Numerical methodsMedium
Approximate ∫01 1/(1 + x³) dx with the trapezoid rule using n = 2 strips. Give T.
f(x) = 1/(1 + x³)[a, b] = [0, 1]n = 2
T =
8.Numerical methodsHard
Use Euler's method with h = 0.1 for y′ = t − y, y(0) = 0.5 to estimate y5 ≈ y(0.5).
y′ = t − yy(0) = 0.5h = 0.1
y5 =
3Solutions
1.Differential equations
Answer: T = 187.1 °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) = 30 + 310·e−0.04t
Substitute t
T(17) = 30 + 310·0.5066 = 187.1 °C
Answer
T = 187.1 °C
2.Differential equations
Answer: t = 173.3 min
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.
Result
m(t) = 100 + (−90)·e−t/250
Solve for t
e−t/250 = (100 − 55)/(100 − 10) = 0.5
t = 250·ln(2) = 173.3 min
Answer
t = 173.3 min
3.Integrals
Answer: I = 1.099
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) = ln x
Evaluate at the limits
I = (ln 6 − ln 2) = ln 3 = 1.099
Answer
I = 1.099
4.Integrals
Answer: I = 0.7183
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
2.718·(1) − 1·2 = 0.7183
Answer
I = 0.7183
5.Derivatives
Answer: f′(2) = 0.8
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 for the logarithm
f′(x) = 2x/(x2 + 1)
Substitute
f′(2) = 4/5 = 0.8
Answer
f′(2) = 0.8
6.Derivatives
Answer: b = −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(2) = −3
f′(x) = 2x − 3 → f′(2) = 1
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = −3 − 1·2 = −5
Answer
b = −5
7.Numerical methods
Answer: T = 0.8194
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 = 1 + 2·0.888889 + 0.5 = 3.277778
Trapezoid rule
T = (0.5/2)·3.277778 = 0.8194
Answer
T = 0.8194
This integral has no elementary antiderivative, which is exactly when a numerical rule is needed.
8.Numerical methods
Answer: y5 = 0.3857
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
y₁ = 0.5 + 0.1·(−0.5) = 0.45
y₂ = 0.45 + 0.1·(−0.35) = 0.415
y₃ = 0.415 + 0.1·(−0.215) = 0.3935
y₄ = 0.3935 + 0.1·(−0.0935) = 0.38415
y₅ = 0.38415 + 0.1·(0.01585) = 0.385735
Graph
y5 = 0.3857
Answer
y5 = 0.3857
The exact value is y(0.5) = 0.409796.
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