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 25 °C. After 16 min its temperature is 17 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). Find the constant k in min⁻¹.
T0 = 4 °CTa = 25 °CT(16) = 17 °C
k = min⁻¹
2.Differential equationsHard
Solve the initial value problem y′ + 3y/x = 3x2 for x > 0, with y(1) = 1. Find y(2).
y′ + 3y/x = 3x2y(1) = 1
y(2) =
3.IntegralsMedium
Evaluate I = ∫210 3/x dx.
f(x) = 3/x[2, 10]
I =
4.IntegralsHard
The region between the line y = x and the parabola y = 0.5x² is rotated about the x-axis. Find the volume V of the solid.
y = xy = 0.5x²
V =
5.DerivativesMedium
Let f(x) = ln(2x2 + 5). Find f′(−2).
f(x) = ln(2x2 + 5)x = −2
f′(−2) =
6.DerivativesHard
The tangent line to f(x) = 0.5x2 − 4x + 3 at x = 2 crosses the y-axis at y = b. Find b.
f(x) = 0.5x2 − 4x + 3x = 2
b =
7.Numerical methodsMedium
Approximate ∫01.5 sin(x²) dx with Simpson's rule using n = 2 strips. Give S.
f(x) = sin(x²)[a, b] = [0, 1.5]n = 2
S =
8.Numerical methodsHard
Take one step of the classical Runge–Kutta method (RK4) with h = 0.1 for y′ = cos t − y, y(0) = −1. Find y₁ ≈ y(0.1).
y′ = cos t − yy(0) = −1h = 0.1
y1 =
3Solutions
1.Differential equations
Answer: k = 0.06032 min⁻¹
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.
Ratio of temperature differences
e−16k = -8/-21
Solve for k
k = ln(-21/-8)/16 = 0.06032 min⁻¹
Answer
k = 0.06032 min⁻¹
2.Differential equations
Answer: y(2) = 4.063
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)′ = 3x5 → x3y = 3x6/6 + C
Initial condition
1 = 3/6 + C → C = 0.5
Evaluate
y(2) = (3·26/6 + 0.5)/23 = 4.063
Answer
y(2) = 4.063
3.Integrals
Answer: I = 4.828
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) = 3·ln x
Evaluate at the limits
I = 3·(ln 10 − ln 2) = 3·ln 5 = 4.828
Answer
I = 4.828
4.Integrals
Answer: V = 3.351
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
1x = 0.5x² → x = 0, x = 2
Washer method
V = π∫₀2 ((1x)² − (0.5x²)²) dx
Integrate
V = π(1·2³/3 − 0.25·2⁵/5) = π·1.067 = 3.351
Answer
V = 3.351
5.Derivatives
Answer: f′(−2) = −0.6154
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) = 4x/(2x2 + 5)
Substitute
f′(−2) = −8/13 = −0.6154
Answer
f′(−2) = −0.6154
6.Derivatives
Answer: b = 1
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) = x − 4 → f′(2) = −2
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = −3 − (−2)·2 = 1
Answer
b = 1
7.Numerical methods
Answer: S = 0.7278
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 = 0.75
Σ wᵢfᵢ = 0 + 4·0.533303 + 0.778073 = 2.911284
Simpson's rule
S = (0.75/3)·2.911284 = 0.7278
Answer
S = 0.7278
This integral has no elementary antiderivative, which is exactly when a numerical rule is needed.
8.Numerical methods
Answer: y1 = −0.8098
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(0.05, −0.9) = 1.89875
k₃ = f(0.05, −0.905062) = 1.903813
k₄ = f(0.1, −0.809619) = 1.804623
Combine
y₁ = −1 + (0.1/6)(2 + 2·1.89875 + 2·1.903813 + 1.804623) = −0.8098
Answer
y1 = −0.8098
ExamProblems · examproblems.com/courses/differential-equations