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Differential Equations

Comprehensive guide to ordinary and partial differential equations, solution methods, and applications in science and engineering.

differential-equations odes pdes modeling systems laplace-transforms series-solutions

Prerequisites

Required:

  • Calculus - All topics including derivatives, integrals, and multivariable calculus basics
  • Algebra - Strong foundation in algebraic manipulation

Helpful:

Overview

Differential equations are mathematical equations that relate functions with their derivatives. They model how quantities change over time or space and are fundamental to describing natural phenomena, from population growth to heat transfer, from mechanical vibrations to electrical circuits.

This comprehensive guide covers ordinary differential equations (ODEs), systems of differential equations, solution methods, and introduces partial differential equations (PDEs) with applications throughout science and engineering.

Introduction to Differential Equations

What is a Differential Equation?

A differential equation is an equation involving an unknown function and its derivatives.

Examples:
dy/dx = 2x                    (first-order ODE)
d²y/dx² + y = 0              (second-order ODE)
∂u/∂t = k ∂²u/∂x²           (partial differential equation)

Components:
• Unknown function: y(x), u(x,t), etc.
• Independent variable(s): x, t, etc.
• Derivatives: dy/dx, d²y/dx², ∂u/∂t, etc.
• Given functions: coefficients, source terms

Classification

By Type

Ordinary Differential Equation (ODE):
• One independent variable
• Example: dy/dx = 3x + 2y

Partial Differential Equation (PDE):  
• Multiple independent variables
• Example: ∂u/∂t = ∂²u/∂x²

By Order

Order = highest derivative in the equation

First-order: dy/dx = f(x, y)
Second-order: d²y/dx² = f(x, y, dy/dx)
nth-order: d^n y/dx^n = f(x, y, dy/dx, ..., d^(n-1)y/dx^(n-1))

By Linearity

Linear: coefficients depend only on independent variable
dy/dx + p(x)y = q(x)

Nonlinear: coefficients depend on y or its derivatives
dy/dx = y²
d²y/dx² + sin(y) = 0

Solutions

General Solution: contains arbitrary constants
Example: y = Ce^x is general solution to dy/dx = y

Particular Solution: specific values for constants
Example: y = 3e^x satisfies dy/dx = y and y(0) = 3

Initial Value Problem (IVP): DE + initial conditions
dy/dx = 2x, y(0) = 1

Boundary Value Problem (BVP): DE + boundary conditions
d²y/dx² + y = 0, y(0) = 0, y(π) = 0

First-Order ODEs

Separable Equations

Form: dy/dx = f(x)g(y)

Method: Separate variables and integrate
dy/g(y) = f(x)dx
∫ dy/g(y) = ∫ f(x)dx

Example: dy/dx = xy
dy/y = x dx
∫ dy/y = ∫ x dx
ln|y| = x²/2 + C
y = Ae^(x²/2) where A = ±e^C

Example: dy/dx = (1 + y²)/(1 + x²)
dy/(1 + y²) = dx/(1 + x²)
∫ dy/(1 + y²) = ∫ dx/(1 + x²)
arctan(y) = arctan(x) + C
y = tan(arctan(x) + C)

Linear First-Order ODEs

Form: dy/dx + P(x)y = Q(x)

Method: Integrating factor μ(x) = e^∫P(x)dx

Steps:
1. Find integrating factor μ(x) = e^∫P(x)dx
2. Multiply equation by μ(x): μ(dy/dx) + μP(x)y = μQ(x)
3. Left side is d/dx[μ(x)y]
4. Integrate: μ(x)y = ∫μ(x)Q(x)dx
5. Solve for y

Example: dy/dx + 2y = e^(-x)
P(x) = 2, Q(x) = e^(-x)
μ(x) = e^∫2dx = e^(2x)

Multiply by μ(x): e^(2x)dy/dx + 2e^(2x)y = e^(2x)e^(-x) = e^x
d/dx[e^(2x)y] = e^x
e^(2x)y = ∫e^x dx = e^x + C
y = e^(-x) + Ce^(-2x)

Exact Equations

Form: M(x,y)dx + N(x,y)dy = 0 where ∂M/∂y = ∂N/∂x

Method: Find function F(x,y) such that ∂F/∂x = M and ∂F/∂y = N

Example: (2xy + 3)dx + (x² + 4y)dy = 0
Check exactness: ∂M/∂y = 2x, ∂N/∂x = 2x ✓

Find F(x,y):
∂F/∂x = 2xy + 3, so F = x²y + 3x + g(y)
∂F/∂y = x² + g'(y) = x² + 4y
Therefore g'(y) = 4y, so g(y) = 2y²

Solution: F(x,y) = x²y + 3x + 2y² = C

Substitution Methods

Homogeneous Equations

Form: dy/dx = f(y/x)

Substitution: v = y/x, so y = vx and dy/dx = v + x(dv/dx)

Example: dy/dx = (y + x)/(y - x) = (y/x + 1)/(y/x - 1)
Let v = y/x: v + x(dv/dx) = (v + 1)/(v - 1)
x(dv/dx) = (v + 1)/(v - 1) - v = 2/(v - 1)
(v - 1)dv = 2dx/x
∫(v - 1)dv = 2∫dx/x
v²/2 - v = 2ln|x| + C
Substitute back: y²/(2x²) - y/x = 2ln|x| + C

Applications of First-Order ODEs

Population Growth

Exponential Growth: dP/dt = kP
Solution: P(t) = P₀e^(kt)

Logistic Growth: dP/dt = kP(1 - P/K)
where K is carrying capacity
Solution: P(t) = K/(1 + Ae^(-kt))

Newton’s Law of Cooling

dT/dt = -k(T - T_ambient)
Solution: T(t) = T_ambient + (T₀ - T_ambient)e^(-kt)

RC Circuits

Kirchhoff's Law: R(dI/dt) + I/C = V₀/R
where I is current, V₀ is applied voltage

Second-Order Linear ODEs

General Form

a(x)d²y/dx² + b(x)dy/dx + c(x)y = f(x)

Homogeneous: f(x) = 0
Non-homogeneous: f(x) ≠ 0

Solution structure: y = y_h + y_p
• y_h: homogeneous solution (general solution to homogeneous equation)
• y_p: particular solution (any solution to non-homogeneous equation)

Constant Coefficient Homogeneous Equations

Form: ay’’ + by’ + cy = 0

Method: Assume solution y = e^(rx)
Characteristic equation: ar² + br + c = 0

Case 1: Two distinct real roots r₁, r₂
y = C₁e^(r₁x) + C₂e^(r₂x)

Case 2: One repeated real root r
y = (C₁ + C₂x)e^(rx)

Case 3: Complex roots r = α ± βi
y = e^(αx)(C₁cos(βx) + C₂sin(βx))

Example: y'' - 3y' + 2y = 0
Characteristic equation: r² - 3r + 2 = 0
(r - 1)(r - 2) = 0, so r₁ = 1, r₂ = 2
General solution: y = C₁e^x + C₂e^(2x)

Example: y'' + 4y' + 4y = 0
Characteristic equation: r² + 4r + 4 = 0
(r + 2)² = 0, so r = -2 (repeated)
General solution: y = (C₁ + C₂x)e^(-2x)

Example: y'' + y = 0
Characteristic equation: r² + 1 = 0
r = ±i = 0 ± 1·i, so α = 0, β = 1
General solution: y = C₁cos(x) + C₂sin(x)

Method of Undetermined Coefficients

For non-homogeneous equations ay’’ + by’ + cy = f(x) where f(x) has specific forms.

Trial solutions based on f(x):

f(x) = polynomial → try polynomial
f(x) = e^(kx) → try Ae^(kx)
f(x) = sin(kx) or cos(kx) → try A sin(kx) + B cos(kx)
f(x) = x^n e^(kx) → try polynomial × e^(kx)

If trial solution matches homogeneous solution, multiply by x.

Example: y'' - 3y' + 2y = e^x
Homogeneous solution: y_h = C₁e^x + C₂e^(2x)
Since e^x appears in y_h, try y_p = Axe^x

y_p' = Ae^x + Axe^x
y_p'' = 2Ae^x + Axe^x

Substitute: (2A + Ax)e^x - 3(A + Ax)e^x + 2Axe^x = e^x
Simplify: (2A - 3A)e^x = e^x
-A = 1, so A = -1

Particular solution: y_p = -xe^x
General solution: y = C₁e^x + C₂e^(2x) - xe^x

Variation of Parameters

General method for finding particular solutions when coefficients are not constant or f(x) doesn’t fit standard forms.

For y'' + P(x)y' + Q(x)y = f(x):

If y_h = C₁y₁(x) + C₂y₂(x), then:
y_p = u₁(x)y₁(x) + u₂(x)y₂(x)

where u₁' and u₂' satisfy:
y₁u₁' + y₂u₂' = 0
y₁'u₁' + y₂'u₂' = f(x)

Solve using Cramer's rule:
u₁' = -y₂f(x)/W, u₂' = y₁f(x)/W
where W = y₁y₂' - y₁'y₂ (Wronskian)

Applications of Second-Order ODEs

Simple Harmonic Motion

mx'' + kx = 0 (mass-spring system)
Solution: x(t) = A cos(ωt) + B sin(ωt) where ω = √(k/m)

Alternative form: x(t) = C cos(ωt + φ)
where C = amplitude, φ = phase angle

Damped Harmonic Motion

mx'' + cx' + kx = 0

Cases:
• Overdamped: c² > 4mk (two real roots)
• Critically damped: c² = 4mk (repeated root)
• Underdamped: c² < 4mk (complex roots)

Underdamped solution: x(t) = e^(-ct/2m)[A cos(ωt) + B sin(ωt)]
where ω = √(4mk - c²)/(2m)

RLC Circuits

LQ'' + RQ' + Q/C = E(t)
where Q is charge, L is inductance, R is resistance, C is capacitance

Systems of Differential Equations

Linear Systems

Matrix form: x' = Ax + f(t)
where x is vector of unknowns, A is coefficient matrix

Example:
x₁' = 2x₁ - x₂
x₂' = x₁ + 2x₂

Matrix form: [x₁'] = [2  -1] [x₁]
             [x₂']   [1   2] [x₂]

Solution by Eigenvalues

For homogeneous system x’ = Ax:

Steps:
1. Find eigenvalues λ of matrix A
2. Find corresponding eigenvectors v
3. Solutions are x = ve^(λt)

Example: A = [1   2]
             [2  -2]

Characteristic equation: det(A - λI) = (1-λ)(-2-λ) - 4 = λ² + λ - 6 = 0
(λ + 3)(λ - 2) = 0, so λ₁ = -3, λ₂ = 2

For λ₁ = -3: (A + 3I)v = 0
[4  2] [v₁] = [0]  → v₁ = [1]
[2  1] [v₂]   [0]      [-2]

For λ₂ = 2: (A - 2I)v = 0  
[-1  2] [v₁] = [0]  → v₂ = [2]
[2  -4] [v₂]   [0]      [1]

General solution: x = C₁[1]e^(-3t) + C₂[2]e^(2t)
                      [-2]         [1]

Phase Portraits

Graphical representation of solution trajectories in the phase plane.

Critical points: equilibrium solutions where x' = 0

Classification by eigenvalues:
• Both λ < 0: stable node (sink)
• Both λ > 0: unstable node (source)
• λ₁ < 0 < λ₂: saddle point
• Complex λ = α ± βi:
  - α < 0: stable spiral
  - α > 0: unstable spiral
  - α = 0: center (neutral)

Laplace Transforms

Definition

L[f(t)] = F(s) = ∫₀^∞ e^(-st)f(t) dt

Common transforms:
L[1] = 1/s
L[t] = 1/s²
L[t^n] = n!/s^(n+1)
L[e^(at)] = 1/(s-a)
L[sin(at)] = a/(s² + a²)
L[cos(at)] = s/(s² + a²)

Properties

Linearity: L[af(t) + bg(t)] = aL[f(t)] + bL[g(t)]
First derivative: L[f'(t)] = sL[f(t)] - f(0)
Second derivative: L[f''(t)] = s²L[f(t)] - sf(0) - f'(0)
Translation: L[e^(at)f(t)] = F(s-a)
Convolution: L[f * g] = F(s)G(s)

Solving ODEs with Laplace Transforms

Example: y'' + 4y = 0, y(0) = 1, y'(0) = 0

Take Laplace transform:
L[y''] + 4L[y] = 0
s²Y(s) - sy(0) - y'(0) + 4Y(s) = 0
s²Y(s) - s + 4Y(s) = 0
Y(s)(s² + 4) = s
Y(s) = s/(s² + 4)

Inverse transform: y(t) = cos(2t)

Introduction to PDEs

Classification

Order: highest partial derivative

Linear second-order PDE:
Auₓₓ + Buₓᵧ + Cuᵧᵧ + Duₓ + Euᵧ + Fu = G

Classification by discriminant Δ = B² - 4AC:
• Δ < 0: Elliptic (e.g., Laplace equation)
• Δ = 0: Parabolic (e.g., heat equation)  
• Δ > 0: Hyperbolic (e.g., wave equation)

Important PDEs

Heat Equation (Parabolic)

∂u/∂t = k ∂²u/∂x²

Physical interpretation: temperature distribution u(x,t)
k = thermal diffusivity

Solution by separation of variables:
u(x,t) = X(x)T(t)
Leads to: X''/X = T'/(kT) = -λ (constant)

Wave Equation (Hyperbolic)

∂²u/∂t² = c² ∂²u/∂x²

Physical interpretation: vibrating string u(x,t)
c = wave speed

General solution: u(x,t) = f(x - ct) + g(x + ct)
(d'Alembert's formula)

Laplace Equation (Elliptic)

∇²u = ∂²u/∂x² + ∂²u/∂y² = 0

Physical interpretation: steady-state heat distribution
Solutions called harmonic functions

In polar coordinates: ∇²u = uᵣᵣ + (1/r)uᵣ + (1/r²)u_θθ = 0

Separation of Variables

Standard method for solving linear PDEs.

Example: Heat equation uₜ = kuₓₓ on [0,L] with u(0,t) = u(L,t) = 0

Assume u(x,t) = X(x)T(t):
X(x)T'(t) = kX''(x)T(t)
T'/T = kX''/X = -λ (constant)

This gives:
T' + λkT = 0  →  T(t) = e^(-λkt)
X'' + (λ/k)X = 0

For boundary conditions X(0) = X(L) = 0:
λₙ = (nπ/L)², Xₙ(x) = sin(nπx/L), n = 1,2,3,...

General solution: u(x,t) = Σ Aₙ sin(nπx/L) e^(-n²π²kt/L²)
where Aₙ determined by initial conditions

Numerical Methods for ODEs

Euler’s Method

Simplest numerical method for solving y’ = f(x,y), y(x₀) = y₀.

Algorithm:
xₙ₊₁ = xₙ + h
yₙ₊₁ = yₙ + hf(xₙ, yₙ)

Example: y' = x + y, y(0) = 1, find y(0.2) with h = 0.1

x₀ = 0, y₀ = 1
y₁ = y₀ + h(x₀ + y₀) = 1 + 0.1(0 + 1) = 1.1

x₁ = 0.1, y₁ = 1.1  
y₂ = y₁ + h(x₁ + y₁) = 1.1 + 0.1(0.1 + 1.1) = 1.22

Therefore y(0.2) ≈ 1.22

Runge-Kutta Methods

More accurate than Euler’s method.

Fourth-order Runge-Kutta (RK4):
k₁ = hf(xₙ, yₙ)
k₂ = hf(xₙ + h/2, yₙ + k₁/2)
k₃ = hf(xₙ + h/2, yₙ + k₂/2)  
k₄ = hf(xₙ + h, yₙ + k₃)

yₙ₊₁ = yₙ + (k₁ + 2k₂ + 2k₃ + k₄)/6

Practice Problems

First-Order ODEs

1. Solve: dy/dx = x/y
2. Solve: dy/dx + 3y = e^(-2x)
3. Solve: (2x + y)dx + (x + 2y)dy = 0
4. Find the solution to: dy/dt = 2y, y(0) = 3

Second-Order ODEs

1. Solve: y'' - 4y' + 3y = 0
2. Solve: y'' + 4y = sin(x)
3. Solve: y'' + 2y' + y = 0, y(0) = 1, y'(0) = 0
4. Find general solution: y'' - 6y' + 9y = e^(3x)

Applications

1. A tank contains 100 gallons of brine with 50 pounds of salt. 
   Water flows in at 5 gal/min and brine flows out at 5 gal/min.
   Find the amount of salt at time t.

2. A mass of 2 kg is attached to a spring with k = 32 N/m.
   Find the equation of motion and period of oscillation.

3. An RC circuit has R = 100Ω, C = 0.01F, and voltage V = 12V.
   Find the current as a function of time if I(0) = 0.

Systems

1. Solve the system: x' = x + 2y, y' = 2x + y

2. Find the critical points and classify them:
   x' = x - y, y' = x + y

Answer Key

First-Order ODEs

1. x dx = y dy → x²/2 = y²/2 + C → x² - y² = C
2. μ(x) = e^(3x), y = e^(-2x)/5 + Ce^(-3x)
3. Check exactness: ∂M/∂y = 1 = ∂N/∂x ✓
   Solution: x² + xy + y² = C
4. y = 3e^(2t)

Second-Order ODEs

1. r² - 4r + 3 = 0 → r = 1, 3
   y = C₁e^x + C₂e^(3x)

2. y_h = C₁cos(2x) + C₂sin(2x)
   Try y_p = A sin(x) → A = 1/3
   y = C₁cos(2x) + C₂sin(2x) + (1/3)sin(x)

3. r² + 2r + 1 = 0 → r = -1 (repeated)
   y = (C₁ + C₂x)e^(-x)
   Apply ICs: y = (1 + 2x)e^(-x)

4. Homogeneous: y_h = (C₁ + C₂x)e^(3x)
   Since e^(3x) in y_h, try y_p = Ax²e^(3x)
   y = (C₁ + C₂x + x²/2)e^(3x)

Applications

1. dS/dt = -S/20, S(0) = 50
   S(t) = 50e^(-t/20) pounds

2. 2x'' + 32x = 0 → x'' + 16x = 0
   Period T = 2π/4 = π/2 seconds

3. RC dI/dt + I = 0, I(0) = 0.12A
   I(t) = 0.12e^(-t) amperes

Systems

1. Eigenvalues: λ = 3, -1
   x = C₁[1]e^(3t) + C₂[ 1]e^(-t)
         [1]        [-1]

2. Critical point: (0,0)
   Eigenvalues: λ = 1 ± i
   Classification: unstable spiral

Quick Reference

Solution Methods

  • Separable: dy/g(y) = f(x)dx
  • Linear first-order: Use integrating factor μ = e^∫P dx
  • Second-order constant coefficient: Characteristic equation
  • Undetermined coefficients: Match form of forcing function
  • Variation of parameters: Use Wronskian

Laplace Transforms

  • L[f’] = sF(s) - f(0)
  • L[f’’] = s²F(s) - sf(0) - f’(0)
  • L[e^(at)f(t)] = F(s-a)

Common Solutions

  • Exponential growth: P = P₀e^(kt)
  • Simple harmonic: x = A cos(ωt) + B sin(ωt)
  • Damped oscillation: x = e^(-ct)[A cos(ωt) + B sin(ωt)]

See Also

Categories:
math
Last updated: January 1, 2023