Second Order Differential Equations
3.7 Cauchy-Euler Equation
The Cauchy-Euler equation, also known as the Euler-Cauchy equation or simply Euler’s equation, is a type of second-order linear differential equation with variable coefficients that appear in many applications in physics and engineering. These equations are particularly noteworthy because they have variable coefficients that are powers of the independent variable.
A second-order Cauchy-Euler equation is generally of the form:
[asciimath]ax^2y''+bxy'+cy=f(x)[/asciimath] (3.7.1)
Here [asciimath]a,\ b,[/asciimath] and [asciimath]c[/asciimath] are constant and [asciimath]f(x)[/asciimath] is a function of the independent variable. The equation is homogeneous if [asciimath]f(x)=0[/asciimath] and inhomogeneous otherwise. For example, [asciimath]-3x^2y''+4xy'+y=cosx[/asciimath] is a Cauchy-Euler equation.
Method to Solve a Homogeneous Cauchy-Euler Equation
To solve a homogeneous Cauchy-Euler Equation 3.7.1,
1. Substitute and Transform: Let [asciimath]y=x^r[/asciimath] and form the characteristic (auxiliary) equation. Thus, [asciimath]y'=rx^(r-1)[/asciimath] , and [asciimath]y''=r(r-1)x^(r-2)[/asciimath]. Substituting these into Equation 3.7.1, we obtain
[asciimath]ar(r-1)x^r+brx^r+cx^r=0[/asciimath] [asciimath]->[/asciimath] [asciimath]x^r(ar(r-1)+br+c)=0[/asciimath]
[asciimath]->[/asciimath] [asciimath]ar(r-1)+br+c=0[/asciimath]
which yields the characteristic equation.
[asciimath]ar^2+(b-a)r+c=0[/asciimath]
2. Solve the Characteristic Equation: Similar to the equations with constant coefficients, we solve the quadratic equation for [asciimath]r[/asciimath], and depending on the nature of the roots, the solution will have different forms.
Case 1: Two Distinct Real Roots [asciimath]r_1[/asciimath] and [asciimath]r_1[/asciimath]
The general solution will be the linear combination of [asciimath]y_1=x^(r_1)[/asciimath] and [asciimath]y_2=x^(r_2)[/asciimath]:
[asciimath]y=c_1 x^(r_1) + c_2 x^(r_2)[/asciimath]
Case 2: Repeated Root [asciimath]r[/asciimath]
The general solution will be the linear combination of [asciimath]y_1=x^r[/asciimath] and [asciimath]y_2=x^rlnx[/asciimath]:
[asciimath]y=c_1 x^r + c_2x^rlnx[/asciimath]
Case 3: Complex Conjugate Roots [asciimath]r=alpha+-betai[/asciimath]
The general solution will be the linear combination of [asciimath]y_1=x^(alpha) cos(beta lnx)[/asciimath] and [asciimath]y_2=x^(alpha) sin(beta lnx)[/asciimath] :
[asciimath]y=x^(alpha)(c_1cos(beta lnx) +c_2sin (betalnx))[/asciimath]
Solve the initial value problem
[asciimath]-x^2y''+7x y'-16y = 0;[/asciimath] [asciimath]\ \ y(1)=-4,\ y'(1)=-1[/asciimath]
Show/Hide Solution
The equation is Cauchy-Euler.
1. So first we find its characteristic polynomial given [asciimath]a=-1[/asciimath], [asciimath]b=7[/asciimath], and [asciimath]c=-16[/asciimath] :
[asciimath]-r^2+8r-16=0[/asciimath]
The equation has a repeated root [asciimath]r=4[/asciimath], which is Case 2.
2. Therefore, the general solution of the equation is
[asciimath]y(x)=c_1 x^4 + c_2x^4lnx[/asciimath]
3. We use the initial values to find [asciimath]c_1[/asciimath] and [asciimath]c_2[/asciimath]:
[asciimath]y(1)=-4[/asciimath]
[asciimath]c_1(1)^4+c_2(1)^4ln(1)=-4[/asciimath] [asciimath]->[/asciimath] [asciimath]c_1=-4[/asciimath]
[asciimath]y'(1)=-1[/asciimath]
[asciimath]4c_1(1)^3+4c_2(1)^3ln(1)+c_2(1)^3=-1[/asciimath] [asciimath]->[/asciimath] [asciimath]c_2=15[/asciimath]
Therefore, the solution to the IVP is
[asciimath]y(x)=-4x^4+15x^4lnx[/asciimath]
Try an Example
For a nonhomogeneous Cauchy-Euler equation, the method of variation of parameters or undetermined coefficients (if applicable) is used.
Section 3.7 Exercises
- Find the general solution of the following equation.
[asciimath]-x^2y''+4x y'-4y = 0[/asciimath]
Show/Hide Answer
[asciimath]y(x)=c_1x^4+c_2x[/asciimath]
- Solve the initial value problem
[asciimath]-2x^2y''-26x y'-70y = 0;[/asciimath] [asciimath]\ \ y(1)=1,\ y'(1)=2[/asciimath]
Show/Hide Answer
[asciimath]y(x)=9/2x^-5-7/2x^-7[/asciimath]
