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Linear Algebra and Differential Equations Unit 9 Review

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9.3 Cauchy-Euler Equations

9.3 Cauchy-Euler Equations

Written by the Fiveable Content Team • Last updated August 2025
Written by the Fiveable Content Team • Last updated August 2025
Linear Algebra and Differential Equations
Unit & Topic Study Guides

Cauchy-Euler equations are a special type of linear differential equations where the variable appears as a power in each term. They pop up in real-world problems like heat conduction in tapered rods and beam vibrations with varying cross-sections.

These equations have a unique structure that allows for specialized solution techniques. By transforming them into linear equations with constant coefficients, we can use the characteristic equation method to solve them, making them a key part of higher-order linear differential equations.

Cauchy-Euler equations

Characteristics and applications

  • Cauchy-Euler equations comprise a special class of linear differential equations with variable coefficients where the variable appears as a power in each term
  • Second-order Cauchy-Euler equation general form ax2y+bxy+cy=f(x)ax^2y'' + bxy' + cy = f(x) with a, b, and c as constants, and f(x) as a function of x or zero
  • Also known as Euler equations or equidimensional equations due to their unique structure
  • Arise in problems involving heat conduction in tapered rods, vibrations of beams with varying cross-sections, and certain economic models
  • Coefficients of the highest-order derivative term and the independent variable x always have the same power
  • Maintain their form under the transformation x=etx = e^t, which enables their solution method
  • Higher-order Cauchy-Euler equations follow a similar pattern with the general form anxny(n)+an1xn1y(n1)+...+a1xy+a0y=f(x)a_nx^ny^{(n)} + a_{n-1}x^{n-1}y^{(n-1)} + ... + a_1xy' + a_0y = f(x)
    • n represents the order of the equation
    • a_n, a_{n-1}, ..., a_1, a_0 are constants
  • Unique structure allows for specialized solution techniques
    • Enables transformation into linear equations with constant coefficients
    • Facilitates the use of characteristic equation method

Examples and applications

  • Heat conduction in tapered rods
    • Models temperature distribution in non-uniform heat-conducting materials
    • Example equation: x2d2Tdx2+xdTdx2T=0x^2\frac{d^2T}{dx^2} + x\frac{dT}{dx} - 2T = 0
      • T represents temperature
      • x represents distance along the rod
  • Vibrations of beams with varying cross-sections
    • Describes the displacement of a non-uniform beam under stress
    • Example equation: x2d2ydx2+3xdydx+y=0x^2\frac{d^2y}{dx^2} + 3x\frac{dy}{dx} + y = 0
      • y represents displacement
      • x represents position along the beam
  • Economic models
    • Used in certain growth models and financial calculations
    • Example equation: x2d2Pdx2+4xdPdx2P=0x^2\frac{d^2P}{dx^2} + 4x\frac{dP}{dx} - 2P = 0
      • P represents price or economic variable
      • x represents time or another economic factor

Transformation of Cauchy-Euler equations

Substitution process

  • Primary substitution used x=etx = e^t or equivalently t=ln(x)t = \ln(x)
  • Substitution leads to dxdt=et\frac{dx}{dt} = e^t and d2xdt2=et\frac{d^2x}{dt^2} = e^t, crucial in the transformation process
  • Apply chain rule to express derivatives with respect to x in terms of derivatives with respect to t
  • For second-order Cauchy-Euler equation, transformation yields:
    • dydx=1xdydt\frac{dy}{dx} = \frac{1}{x}\frac{dy}{dt}
    • d2ydx2=1x2(d2ydt2dydt)\frac{d^2y}{dx^2} = \frac{1}{x^2}(\frac{d^2y}{dt^2} - \frac{dy}{dt})
  • After substitution and simplification, resulting equation has constant coefficients in terms of t
  • Boundary of original equation (typically x > 0) transforms to <t<-\infty < t < \infty in new equation
  • Higher-order Cauchy-Euler equations follow similar transformation pattern
    • Involves more complex applications of the chain rule
    • Results in higher-order linear equations with constant coefficients
Characteristics and applications, Cauchy–Euler equation - Wikipedia

Examples of transformation

  • Example 1: Transform x2y+3xy4y=0x^2y'' + 3xy' - 4y = 0
    • Substitute x=etx = e^t and apply chain rule
    • Resulting equation: d2ydt2+2dydt4y=0\frac{d^2y}{dt^2} + 2\frac{dy}{dt} - 4y = 0
  • Example 2: Transform x3y+2x2yxy+y=0x^3y''' + 2x^2y'' - xy' + y = 0
    • Substitute x=etx = e^t and apply chain rule
    • Resulting equation: d3ydt3+3d2ydt2+2dydt+y=0\frac{d^3y}{dt^3} + 3\frac{d^2y}{dt^2} + 2\frac{dy}{dt} + y = 0

Solving Cauchy-Euler equations

Characteristic equation method

  • After transformation, equation becomes linear differential equation with constant coefficients
  • Form characteristic equation by substituting y=erty = e^{rt} into transformed equation, where r is a constant
  • Roots of characteristic equation determine form of general solution in terms of t
  • For real and distinct roots, solution is linear combination of erite^{r_it}, where r_i are the roots
  • For repeated real roots, solution includes terms of form tkertt^k e^{rt}, where k ranges from 0 to (multiplicity - 1)
  • For complex conjugate roots a ± bi, solution includes terms of form eat(c1cos(bt)+c2sin(bt))e^{at}(c_1\cos(bt) + c_2\sin(bt))
  • Find particular solution for non-homogeneous equations using methods (undetermined coefficients, variation of parameters)
  • Back-substitute t=ln(x)t = \ln(x) to express solution in terms of original variable x

Solution examples

  • Example 1: Solve x2y+3xy4y=0x^2y'' + 3xy' - 4y = 0
    • Transformed equation: d2ydt2+2dydt4y=0\frac{d^2y}{dt^2} + 2\frac{dy}{dt} - 4y = 0
    • Characteristic equation: r2+2r4=0r^2 + 2r - 4 = 0
    • Roots: r=3r = -3 or r=1r = 1
    • General solution: y=c1e3t+c2ety = c_1e^{-3t} + c_2e^t
    • Back-substitute: y=c1x3+c2xy = c_1x^{-3} + c_2x
  • Example 2: Solve x2y+5xy+4y=0x^2y'' + 5xy' + 4y = 0
    • Transformed equation: d2ydt2+4dydt+4y=0\frac{d^2y}{dt^2} + 4\frac{dy}{dt} + 4y = 0
    • Characteristic equation: r2+4r+4=0r^2 + 4r + 4 = 0
    • Repeated root: r=2r = -2
    • General solution: y=(c1+c2t)e2ty = (c_1 + c_2t)e^{-2t}
    • Back-substitute: y=(c1+c2ln(x))x2y = (c_1 + c_2\ln(x))x^{-2}
Characteristics and applications, EulerEquations | Wolfram Function Repository

General solutions vs initial conditions

General solution properties

  • General solution expressed as linear combination of fundamental solutions obtained from characteristic equation method
  • For second-order equation, general solution takes form y=c1y1(x)+c2y2(x)y = c_1y_1(x) + c_2y_2(x), where y_1(x) and y_2(x) are linearly independent solutions
  • Fundamental solutions for Cauchy-Euler equations often involve terms:
    • xrx^r
    • xrln(x)x^r\ln(x)
    • xa(c1cos(bln(x))+c2sin(bln(x)))x^a(c_1\cos(b\ln(x)) + c_2\sin(b\ln(x)))
  • Higher-order Cauchy-Euler equations follow similar pattern with more terms in general solution

Applying initial conditions

  • Find particular solution by applying initial conditions to general solution and its derivatives at specified point (typically x = 1 for convenience)
  • Form system of linear equations using initial conditions to solve for constants c_1, c_2, etc.
  • Uniqueness theorem for linear differential equations ensures unique solution exists for well-posed initial value problems
  • Higher-order Cauchy-Euler equations involve more constants and initial conditions
  • Verify final solution by substituting back into original Cauchy-Euler equation
    • Check satisfaction of both equation and initial conditions

Examples with initial conditions

  • Example 1: Solve x2y+3xy4y=0x^2y'' + 3xy' - 4y = 0 with y(1) = 2 and y'(1) = 3
    • General solution: y=c1x3+c2xy = c_1x^{-3} + c_2x
    • Apply initial conditions:
      • 2=c1+c22 = c_1 + c_2
      • 3=3c1+c23 = -3c_1 + c_2
    • Solve system of equations:
      • c1=1c_1 = -1, c2=3c_2 = 3
    • Particular solution: y=x3+3xy = -x^{-3} + 3x
  • Example 2: Solve x2y+5xy+4y=0x^2y'' + 5xy' + 4y = 0 with y(1) = 1 and y'(1) = 0
    • General solution: y=(c1+c2ln(x))x2y = (c_1 + c_2\ln(x))x^{-2}
    • Apply initial conditions:
      • 1=c11 = c_1
      • 0=c22c10 = c_2 - 2c_1
    • Solve system of equations:
      • c1=1c_1 = 1, c2=2c_2 = 2
    • Particular solution: y=(1+2ln(x))x2y = (1 + 2\ln(x))x^{-2}
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