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 27 °C. After 22 min its temperature is 21 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). Find the constant k in min⁻¹.
T0 = 4 °CTa = 27 °CT(22) = 21 °C
k = min⁻¹
2.Differential equationsHard
Solve the initial value problem y′ + 2y/x = 2x2 for x > 0, with y(1) = 0. Find y(2).
y′ + 2y/x = 2x2y(1) = 0
y(2) =
3.IntegralsMedium
Evaluate I = ∫02 2x·(x² + 1)1 dx.
f(x) = 2x·(x² + 1)1[0, 2]
I =
4.IntegralsHard
Evaluate I = ∫₀2.5 x²·eˣ dx.
∫₀2.5 x²eˣ dx
I =
5.DerivativesMedium
Let f(x) = x2·e0.5x. Find f′(2).
f(x) = x2·e0.5xx = 2
f′(2) =
6.DerivativesHard
Let f(x) = x3 + 1.5x2 − 6x − 1. Find the x-coordinate x of the local maximum.
f(x) = x3 + 1.5x2 − 6x − 1
x =
7.Numerical methodsMedium
Approximate ∫01 x³ dx with the trapezoid rule, n = 2, and compare it with the exact value. Find the absolute error |E| = |T − I|.
f(x) = x³[a, b] = [0, 1]n = 2
|E| =
8.Numerical methodsHard
Use Euler's method with h = 0.5 for y′ = y − t², y(0) = 0.5 to estimate y5 ≈ y(2.5).
y′ = y − t²y(0) = 0.5h = 0.5
y5 =
3Solutions
1.Differential equations
Answer: k = 0.06108 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−22k = -6/-23
Solve for k
k = ln(-23/-6)/22 = 0.06108 min⁻¹
Answer
k = 0.06108 min⁻¹
2.Differential equations
Answer: y(2) = 3.1
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.
Integrating factor
μ = e∫2/x dx = x2
Integrate
(x2y)′ = 2x4 → x2y = 2x5/5 + C
Initial condition
0 = 2/5 + C → C = −0.4
Evaluate
y(2) = (2·25/5 + (−0.4))/22 = 3.1
Answer
y(2) = 3.1
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: I = 37.59
Integration by parts: ∫u·dv = u·v − ∫v·du, from the product rule. Choose u as the factor that gets simpler when you differentiate it (a power of x, or ln x), and dv as the factor you can integrate (eˣ, sin x, cos x, or x when u = ln x). For a definite integral, evaluate u·v at both limits: ∫ₐᵇ u dv = [u·v]ₐᵇ − ∫ₐᵇ v du. A factor x² needs two rounds.
First round
∫x²eˣ dx = x²eˣ − ∫2x eˣ dx
Second round
∫2x eˣ dx = 2x eˣ − 2eˣ
∫x²eˣ dx = eˣ(x² − 2x + 2)
Limits
12.18·(3.25) − 1·2 = 37.59
Answer
I = 37.59
5.Derivatives
Answer: f′(2) = 16.31
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′ = e0.5x·(2x + 0.5·x2)
Substitute
f′(2) = 2.718·(4 + 2) = 16.31
Answer
f′(2) = 16.31
6.Derivatives
Answer: x = −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.
Derivative
f′(x) = 3x2 + 3x − 6 = 3·(x2 + x − 2)
Solve f′(x) = 0
x = −2 or x = 1
Which one is which
f″(x) = 6x + 3: negative at a maximum, positive at a minimum.
f″(−2) = −9, f″(1) = 9
x = −2
Answer
x = −2
7.Numerical methods
Answer: |E| = 0.0625
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.
Exact value
I = 0.25
Trapezoid rule
T = (0.5/2)·(0 + 2·0.125 + 1) = 0.3125
Error
|E| = |0.3125 − 0.25| = 0.0625
Answer
|E| = 0.0625
Relative error: 25%.
8.Numerical methods
Answer: y5 = −1.438
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
y₁ = 0.5 + 0.5·(0.5) = 0.75
y₂ = 0.75 + 0.5·(0.5) = 1
y₃ = 1 + 0.5·(0) = 1
y₄ = 1 + 0.5·(−1.25) = 0.375
y₅ = 0.375 + 0.5·(−3.625) = −1.4375
Graph
y5 = −1.438
Answer
y5 = −1.438
The exact value is y(2.5) = −5.023741.
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