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 25 °C. After 58 min its temperature is 41 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). How long after the start, t, does it take to reach 31 °C?
T0 = 320 °CTa = 25 °CT(58) = 41 °CT = 31 °C
t = min
2.Differential equationsHard
Solve the initial value problem y′ + 3y/x = 6 for x > 0, with y(1) = −2. Find y(3).
y′ + 3y/x = 6y(1) = −2
y(3) =
3.IntegralsMedium
The region under y = x from x = 0 to x = 4 is rotated about the x-axis. Find the volume V of the solid (a cone).
y = x0 ≤ x ≤ 4
V =
4.IntegralsHard
The region between the line y = 6x and the parabola y = 3x² is rotated about the x-axis. Find the volume V of the solid.
y = 6xy = 3x²
V =
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
The tangent line to f(x) = 2x2 + x + 4 at x = 1 crosses the y-axis at y = b. Find b.
f(x) = 2x2 + x + 4x = 1
b =
7.Numerical methodsMedium
A vehicle's speed is measured every 1 s (table). Estimate the distance s travelled from t = 0 to t = 5 s with the trapezoid rule.
h = 1 sn = 5
s = m
8.Numerical methodsHard
Take one step of the classical Runge–Kutta method (RK4) with h = 0.1 for y′ = cos t − y, y(0) = 2. Find y₁ ≈ y(0.1).
y′ = cos t − yy(0) = 2h = 0.1
y1 =
3Solutions
1.Differential equations
Answer: t = 77.52 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.
Cooling constant
k = ln(295/16)/58 = 0.05025 min⁻¹
Time to reach the target
t = ln(295/6)/0.05025 = 77.52 min
Answer
t = 77.52 min
2.Differential equations
Answer: y(3) = 4.37
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 = e∫3/x dx = x3
Integrate
(x3y)′ = 6x3 → x3y = 6x4/4 + C
Initial condition
−2 = 6/4 + C → C = −3.5
Evaluate
y(3) = (6·34/4 + (−3.5))/33 = 4.37
Answer
y(3) = 4.37
3.Integrals
Answer: V = 67.02
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 1x² dx
Integrate
V = π·1·4³/3 = 67.02
Answer
V = 67.02
Check: a cone of radius 4 and height 4 has (1/3)πr²h = 67.02.
4.Integrals
Answer: V = 120.6
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.
Intersection points
6x = 3x² → x = 0, x = 2
Washer method
V = π∫₀2 ((6x)² − (3x²)²) dx
Integrate
V = π(36·2³/3 − 9·2⁵/5) = π·38.4 = 120.6
Answer
V = 120.6
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: b = 2
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) = 7
f′(x) = 4x + 1 → f′(1) = 5
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = 7 − 5·1 = 2
Answer
b = 2
7.Numerical methods
Answer: s = 65.5 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
10 + 2·12 + 2·13 + 2·11 + 2·17 + 15 = 131
Trapezoid rule
s = (1/2)·131 = 65.5 m
Answer
s = 65.5 m
8.Numerical methods
Answer: y1 = 1.905
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, 2) = −1
k₂ = f(0.05, 1.95) = −0.95125
k₃ = f(0.05, 1.952438) = −0.953687
k₄ = f(0.1, 1.904631) = −0.909627
Combine
y₁ = 2 + (0.1/6)(−1 + 2·−0.95125 + 2·−0.953687 − 0.909627) = 1.905
Answer
y1 = 1.905
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