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 method to solve them, making them a key part of higher-order linear differential equations.
Cauchy-Euler equations
Characteristics and applications
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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 general form ax2y′′+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=et, which enables their solution method
Higher-order Cauchy-Euler equations follow a similar pattern with the general form anxny(n)+an−1xn−1y(n−1)+...+a1xy′+a0y=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: x2dx2d2T+xdxdT−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: x2dx2d2y+3xdxdy+y=0
y represents displacement
x represents position along the beam
Economic models
Used in certain growth models and financial calculations
Example equation: x2dx2d2P+4xdxdP−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=et or equivalently t=ln(x)
Substitution leads to dtdx=et and dt2d2x=et, 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:
dxdy=x1dtdy
dx2d2y=x21(dt2d2y−dtdy)
After substitution and simplification, resulting equation has constant coefficients in terms of t
Boundary of original equation (typically x > 0) transforms to −∞<t<∞ 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
Examples of transformation
Example 1: Transform x2y′′+3xy′−4y=0
Substitute x=et and apply chain rule
Resulting equation: dt2d2y+2dtdy−4y=0
Example 2: Transform x3y′′′+2x2y′′−xy′+y=0
Substitute x=et and apply chain rule
Resulting equation: dt3d3y+3dt2d2y+2dtdy+y=0
Solving Cauchy-Euler equations
Characteristic equation method
After transformation, equation becomes with constant coefficients
Form characteristic equation by substituting y=ert into transformed equation, where r is a constant
Roots of characteristic equation determine form of in terms of t
For real and distinct roots, solution is linear combination of erit, where r_i are the roots
For repeated real roots, solution includes terms of form tkert, where k ranges from 0 to (multiplicity - 1)
For complex conjugate roots a ± bi, solution includes terms of form eat(c1cos(bt)+c2sin(bt))
Find for non-homogeneous equations using methods (undetermined coefficients, variation of parameters)
Back-substitute t=ln(x) to express solution in terms of original variable x
Solution examples
Example 1: Solve x2y′′+3xy′−4y=0
Transformed equation: dt2d2y+2dtdy−4y=0
Characteristic equation: r2+2r−4=0
Roots: r=−3 or r=1
General solution: y=c1e−3t+c2et
Back-substitute: y=c1x−3+c2x
Example 2: Solve x2y′′+5xy′+4y=0
Transformed equation: dt2d2y+4dtdy+4y=0
Characteristic equation: r2+4r+4=0
Repeated root: r=−2
General solution: y=(c1+c2t)e−2t
Back-substitute: y=(c1+c2ln(x))x−2
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), where y_1(x) and y_2(x) are linearly independent solutions
Fundamental solutions for Cauchy-Euler equations often involve terms:
xr
xrln(x)
xa(c1cos(bln(x))+c2sin(bln(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′′+3xy′−4y=0 with y(1) = 2 and y'(1) = 3
General solution: y=c1x−3+c2x
Apply initial conditions:
2=c1+c2
3=−3c1+c2
Solve system of equations:
c1=−1, c2=3
Particular solution: y=−x−3+3x
Example 2: Solve x2y′′+5xy′+4y=0 with y(1) = 1 and y'(1) = 0