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 3 °C is taken out of the fridge into a room at 27 °C. After 30 min its temperature is 17 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). How long after the start, t, does it take to reach 9 °C?
T0 = 3 °CTa = 27 °CT(30) = 17 °CT = 9 °C
t = min
2.Differential equationsHard
Solve the initial value problem y′ + 3y/x = x2 for x > 0, with y(1) = 5. Find y(3).
y′ + 3y/x = x2y(1) = 5
y(3) =
3.IntegralsMedium
Evaluate I = ∫02 2x·(x² + 1)1 dx.
f(x) = 2x·(x² + 1)1[0, 2]
I =
4.IntegralsHard
The region between the line y = x and the parabola y = x² is rotated about the x-axis. Find the volume V of the solid.
y = xy = x²
V =
5.DerivativesMedium
Let f(x) = x·sin(3x), with x in radians. Find f′(1.5).
f(x) = x·sin(3x)x = 1.5 rad
f′(1.5) =
6.DerivativesHard
Let f(x) = x3 − 7.5x2 + 12x + 5. Find the x-coordinate x of the local maximum.
f(x) = x3 − 7.5x2 + 12x + 5
x =
7.Numerical methodsMedium
Approximate ∫02 √(1 + x³) dx with Simpson's rule using n = 2 strips. Give S.
f(x) = √(1 + x³)[a, b] = [0, 2]n = 2
S =
8.Numerical methodsHard
Solve y′ = y − t², y(0) = 3 with Euler's method and step h = 0.1 up to t = 0.5. The exact solution is y = t² + 2t + 2 + (y₀ − 2)eᵗ. Find the global error |E| = |y(0.5) − y5|.
y′ = y − t²y(0) = 3h = 0.1
|E| =
3Solutions
1.Differential equations
Answer: t = 9.858 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(-24/-10)/30 = 0.02918 min⁻¹
Time to reach the target
t = ln(-24/(−18))/0.02918 = 9.858 min
Answer
t = 9.858 min
2.Differential equations
Answer: y(3) = 4.679
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)′ = x5 → x3y = 1x6/6 + C
Initial condition
5 = 1/6 + C → C = 4.833
Evaluate
y(3) = (1·36/6 + 4.833)/33 = 4.679
Answer
y(3) = 4.679
3.Integrals
Answer: I = 12
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.
Substitution
u = x² + 1, du = 2x dx
I = ∫15 u1 du
Integrate and evaluate
I = [u2/2]15 = (25 − 1)/2 = 12
Answer
I = 12
4.Integrals
Answer: V = 0.4189
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 = 1x² → x = 0, x = 1
Washer method
V = π∫₀1 ((1x)² − (1x²)²) dx
Integrate
V = π(1·1³/3 − 1·1⁵/5) = π·0.1333 = 0.4189
Answer
V = 0.4189
5.Derivatives
Answer: f′(1.5) = −1.926
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(3x) + 3x·cos(3x)
Substitute
f′(1.5) = −0.9775 + 4.5·(−0.2108) = −1.926
Answer
f′(1.5) = −1.926
6.Derivatives
Answer: x = 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.
Derivative
f′(x) = 3x2 − 15x + 12 = 3·(x2 − 5x + 4)
Solve f′(x) = 0
x = 1 or x = 4
Which one is which
f″(x) = 6x − 15: negative at a maximum, positive at a minimum.
f″(1) = −9, f″(4) = 9
x = 1
Answer
x = 1
7.Numerical methods
Answer: S = 3.219
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 = 1
Σ wᵢfᵢ = 1 + 4·1.414214 + 3 = 9.656854
Simpson's rule
S = (1/3)·9.656854 = 3.219
Answer
S = 3.219
This integral has no elementary antiderivative, which is exactly when a numerical rule is needed.
8.Numerical methods
Answer: |E| = 0.09926
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 = 4.799459
Exact value
y(0.5) = 4.898721
Global error
|E| = |4.898721 − 4.799459| = 0.09926
Graph
Answer
|E| = 0.09926
Halving h roughly halves the error of Euler's method.
ExamProblems · examproblems.com/courses/differential-equations