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 5 °C is taken out of the fridge into a room at 30 °C. After 18 min its temperature is 24 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). How long after the start, t, does it take to reach 27 °C?
T0 = 5 °CTa = 30 °CT(18) = 24 °CT = 27 °C
t = min
2.Differential equationsHard
Solve the initial value problem y′ + y/x = 5 for x > 0, with y(1) = −3. Find y(3).
y′ + y/x = 5y(1) = −3
y(3) =
3.IntegralsMedium
Evaluate I = ∫01 e2x dx.
f(x) = e2x[0, 1]
I =
4.IntegralsHard
Find the area A of the region enclosed by y = 2x2 and y = 2x.
y = 2x2y = 2x
A =
5.DerivativesMedium
Let f(x) = x2·e−0.5x. Find f′(1.5).
f(x) = x2·e−0.5xx = 1.5
f′(1.5) =
6.DerivativesHard
Let f(x) = x3 + 6x2 + 4. Find the x-coordinate x of the local maximum.
f(x) = x3 + 6x2 + 4
x =
7.Numerical methodsMedium
A vehicle's speed is measured every 2 s (table). Estimate the distance s travelled from t = 0 to t = 8 s with Simpson's rule.
h = 2 sn = 4
s = m
8.Numerical methodsHard
Take one step of the classical Runge–Kutta method (RK4) with h = 0.5 for y′ = t² + y², y(0) = 1. Find y₁ ≈ y(0.5).
y′ = t² + y²y(0) = 1h = 0.5
y1 =
3Solutions
1.Differential equations
Answer: t = 26.74 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(-25/-6)/18 = 0.07928 min⁻¹
Time to reach the target
t = ln(-25/(−3))/0.07928 = 26.74 min
Answer
t = 26.74 min
2.Differential equations
Answer: y(3) = 5.667
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∫1/x dx = x1
Integrate
(x1y)′ = 5x → x1y = 5x2/2 + C
Initial condition
−3 = 5/2 + C → C = −5.5
Evaluate
y(3) = (5·32/2 + (−5.5))/31 = 5.667
Answer
y(3) = 5.667
3.Integrals
Answer: I = 3.195
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) = e2x/2
Evaluate at the limits
I = (e2 − 1)/2 = (7.389 − 1)/2 = 3.195
Answer
I = 3.195
4.Integrals
Answer: A = 0.3333
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.
Intersection points
2x2 = 2x → x = 0, x = 1
Integrate upper minus lower
A = [x2 − 0.6667x3]01 = 0.3333
Answer
A = 0.3333
5.Derivatives
Answer: f′(1.5) = 0.8857
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 rule
f′(x) = u′v + uv′ = e−0.5x·(2x + (−0.5)·x2)
Substitute
f′(1.5) = 0.4724·(3 + (−1.125)) = 0.8857
Answer
f′(1.5) = 0.8857
6.Derivatives
Answer: x = −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.
Derivative
f′(x) = 3x2 + 12x = 3·(x2 + 4x)
Solve f′(x) = 0
x = −4 or x = 0
Which one is which
f″(x) = 6x + 12: negative at a maximum, positive at a minimum.
f″(−4) = −12, f″(0) = 12
x = −4
Answer
x = −4
7.Numerical methods
Answer: s = 108.7 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
5 + 4·8 + 2·14 + 4·20 + 18 = 163
Simpson's rule
s = (2/3)·163 = 108.7 m
Answer
s = 108.7 m
8.Numerical methods
Answer: y1 = 2.055
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) = 1
k₂ = f(0.25, 1.25) = 1.625
k₃ = f(0.25, 1.40625) = 2.040039
k₄ = f(0.5, 2.02002) = 4.330479
Combine
y₁ = 1 + (0.5/6)(1 + 2·1.625 + 2·2.040039 + 4.330479) = 2.055
Answer
y1 = 2.055
ExamProblems · examproblems.com/courses/differential-equations