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
Solve y″ + 8y′ + 15y = 0 with y(0) = 1 and y′(0) = 4. Find y(1.5).
y″ + 8y′ + 15y = 0y(0) = 1y′(0) = 4
y(1.5) =
2.Differential equationsHard
Solve y′ + 2y = 2e−t with y(0) = 4. Find y(1.5).
y′ + 2y = 2e−ty(0) = 4
y(1.5) =
3.IntegralsMedium
Evaluate I = ∫01 2x·(x² + 1)1 dx.
f(x) = 2x·(x² + 1)1[0, 1]
I =
4.IntegralsHard
Find the area A of the region enclosed by y = 0.5x2 and y = 2x.
y = 0.5x2y = 2x
A =
5.DerivativesMedium
Let f(x) = ln(x2 + 2). Find f′(3).
f(x) = ln(x2 + 2)x = 3
f′(3) =
6.DerivativesHard
The tangent line to f(x) = −x2 − 4x + 2 at x = 2 crosses the y-axis at y = b. Find b.
f(x) = −x2 − 4x + 2x = 2
b =
7.Numerical methodsMedium
Approximate ∫02 e−x² dx with Simpson's rule using n = 2 strips. Give S.
f(x) = e−x²[a, b] = [0, 2]n = 2
S =
8.Numerical methodsHard
Take one step of the classical Runge–Kutta method (RK4) with h = 0.1 for y′ = 1 + sin(ty), y(0) = 0.5. Find y₁ ≈ y(0.1).
y′ = 1 + sin(ty)y(0) = 0.5h = 0.1
y1 =
3Solutions
1.Differential equations
Answer: y(1.5) = 0.04805
Second-order linear equations: for y″ + b·y′ + c·y = 0 try y = ert: this gives the characteristic equation r² + b·r + c = 0. Two real roots r₁ ≠ r₂ give y = A·er₁t + B·er₂t (overdamped); a double root r gives y = (A + Bt)·ert (critically damped); complex roots −α ± jω give y = e−αt(A cos ωt + B sin ωt) (underdamped, oscillating). The initial values y(0) and y′(0) fix A and B. Written as y″ + 2ζωny′ + ωn²y = 0, ωn is the natural frequency and ζ the damping ratio: ζ < 1 oscillates. With a constant input, y″ + by′ + cy = c·K, the solution is the steady value K plus the free response.
Characteristic roots
(r + 3)(r + 5) = 0 → r = −3, −5
Constants from the initial values
A + B = 1, −3A − 5B = 4
A = 4.5, B = −3.5
Evaluate
y(1.5) = 4.5·e−4.5 + (−3.5)·e−7.5 = 0.04805
Answer
y(1.5) = 0.04805
2.Differential equations
Answer: y(1.5) = 0.5458
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 = e2t
(e2ty)′ = 2et
General solution
y = 2/1·e−t + C·e−2t = 2·e−t + C·e−2t
Initial condition
4 = 2 + C → C = 2
Evaluate
y(1.5) = 2·0.2231 + 2·0.04979 = 0.5458
Answer
y(1.5) = 0.5458
3.Integrals
Answer: I = 1.5
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 = ∫12 u1 du
Integrate and evaluate
I = [u2/2]12 = (4 − 1)/2 = 1.5
Answer
I = 1.5
4.Integrals
Answer: A = 5.333
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
0.5x2 = 2x → x = 0, x = 4
Integrate upper minus lower
A = [x2 − 0.1667x3]04 = 5.333
Answer
A = 5.333
5.Derivatives
Answer: f′(3) = 0.5455
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) = 2x/(x2 + 2)
Substitute
f′(3) = 6/11 = 0.5455
Answer
f′(3) = 0.5455
6.Derivatives
Answer: b = 6
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) = −10
f′(x) = −2x − 4 → f′(2) = −8
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = −10 − (−8)·2 = 6
Answer
b = 6
7.Numerical methods
Answer: S = 0.8299
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 + 4·0.367879 + 0.0183156 = 2.489833
Simpson's rule
S = (1/3)·2.489833 = 0.8299
Answer
S = 0.8299
This integral has no elementary antiderivative, which is exactly when a numerical rule is needed.
8.Numerical methods
Answer: y1 = 0.6028
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, 0.5) = 1
k₂ = f(0.05, 0.55) = 1.027497
k₃ = f(0.05, 0.551375) = 1.027565
k₄ = f(0.1, 0.602757) = 1.060239
Combine
y₁ = 0.5 + (0.1/6)(1 + 2·1.027497 + 2·1.027565 + 1.060239) = 0.6028
Answer
y1 = 0.6028
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