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 26 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.03 min⁻¹, find its temperature T after 19 min.
T0 = 4 °CTa = 26 °Ck = 0.03 min⁻¹t = 19 min
T = °C
2.Differential equationsHard
Solve y′ + 0.5y = 3e−3t with y(0) = 5. Find y(1).
y′ + 0.5y = 3e−3ty(0) = 5
y(1) =
3.IntegralsMedium
Evaluate I = ∫26 2/x dx.
f(x) = 2/x[2, 6]
I =
4.IntegralsHard
The region between the line y = 4x and the parabola y = x² is rotated about the x-axis. Find the volume V of the solid.
y = 4xy = x²
V =
5.DerivativesMedium
Let f(x) = x·sin(x), with x in radians. Find f′(1.5).
f(x) = x·sin(x)x = 1.5 rad
f′(1.5) =
6.DerivativesHard
The tangent line to f(x) = x2 − 4x + 5 at x = −1 crosses the y-axis at y = b. Find b.
f(x) = x2 − 4x + 5x = −1
b =
7.Numerical methodsMedium
Approximate ∫02 1/√(1 + x⁴) dx with Simpson's rule using n = 6 strips. Give S.
f(x) = 1/√(1 + x⁴)[a, b] = [0, 2]n = 6
S =
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)e−t. Find the global error |E| = |y(1) − y5|.
y′ = t − yy(0) = 1h = 0.2
|E| =
3Solutions
1.Differential equations
Answer: T = 13.56 °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) = 26 + (−22)·e−0.03t
Substitute t
T(19) = 26 + (−22)·0.5655 = 13.56 °C
Answer
T = 13.56 °C
2.Differential equations
Answer: y(1) = 3.701
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)′ = 3e−2.5t
General solution
y = 3/(−2.5)·e−3t + C·e−0.5t = −1.2·e−3t + C·e−0.5t
Initial condition
5 = −1.2 + C → C = 6.2
Evaluate
y(1) = −1.2·0.04979 + 6.2·0.6065 = 3.701
Answer
y(1) = 3.701
3.Integrals
Answer: I = 2.197
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) = 2·ln x
Evaluate at the limits
I = 2·(ln 6 − ln 2) = 2·ln 3 = 2.197
Answer
I = 2.197
4.Integrals
Answer: V = 428.9
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
4x = 1x² → x = 0, x = 4
Washer method
V = π∫₀4 ((4x)² − (1x²)²) dx
Integrate
V = π(16·4³/3 − 1·4⁵/5) = π·136.5 = 428.9
Answer
V = 428.9
5.Derivatives
Answer: f′(1.5) = 1.104
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′(1.5) = 0.9975 + 1.5·0.07074 = 1.104
Answer
f′(1.5) = 1.104
6.Derivatives
Answer: b = 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.
Point and slope
f(−1) = 10
f′(x) = 2x − 4 → f′(−1) = −6
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = 10 − (−6)·(−1) = 4
Answer
b = 4
7.Numerical methods
Answer: S = 1.357
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 = 0.3333
Σ wᵢfᵢ = 1 + 4·0.993884 + 2·0.913812 + 4·0.707107 + 2·0.490261 + 4·0.338719 + 0.242536 = 12.209521
Simpson's rule
S = (0.3333/3)·12.209521 = 1.357
Answer
S = 1.357
This integral has no elementary antiderivative, which is exactly when a numerical rule is needed.
8.Numerical methods
Answer: |E| = 0.0804
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 = 0.65536
Exact value
y(1) = 0.735759
Global error
|E| = |0.735759 − 0.65536| = 0.0804
Graph
Answer
|E| = 0.0804
Halving h roughly halves the error of Euler's method.
ExamProblems · examproblems.com/courses/differential-equations