Cauchy Euler Equation

The Cauchy–Euler Equation provides a powerful method for solving linear ODEs with variable coefficients.

This equation appears naturally in many physical and engineering problems, especially those solved using Fourier’s or separation-of-variables techniques in partial differential equations (PDEs).

Cauchy-Euler equation, also known as the Euler-Cauchy equation, is a type of ordinary differential equation (ODE) with variable coefficients that are powers of the independent variable.

It’s a linear homogeneous differential equation and is notable for being solvable in terms of power functions. Solutions to the Cauchy-Euler equation can often be found using methods such as substitution, reduction of order, or power series.

Cauchy-Euler Equation General Form

The differential equation,

\bold{a_nx^ny^{(n)}+a_{n-1}x^{n-1}y^{n-1}+...+a_0y}

Where

yⁿ represents the n^th derivative of y with respect to x, a_i, i = 0, . . . , n are constants and a_n ≠ 0.

Cauchy-Euler Equation Examples

Here are some examples of Cauchy-Euler equations:

x²y'' + y = 0
x²y'' + 4y = 0
x²y'' -3xy' - 7y = 0
x³y'' - 2x²y' + xy = 0

Second Order Cauchy-Euler Equation

The most common Cauchy–Euler equation is the second-order equation, appearing in several physics and engineering applications, such as when solving Laplace's equation in polar coordinates. It is given by the equation:

\bold{a_2x^{2}{\frac {d^{2}y}{dx^{2}}}+a_1x{\frac {dy}{dx}}+a_0y=0}

How to Solve the Cauchy-Euler Differential Equation?

Cauchy-Euler differential equations require a few systematic steps to solve, as explained below:

Identification: First, you must verify that the differential equation is of the Cauchy-Euler formax^2y’’ + bxy’ + cy = 0, where the coefficients are powers of x that match the order of the derivative they multiply.

Assumption: You then assume a solution of the form y = x^r, where r is a constant that will be determined.

Calculate Derivatives: You compute the derivatives of the assumed solution y’ and y’’.

Substitution: You have to substitute y, y’, and y’’ back into the original differential equation.

Solve for r: After substitution, you'll get a polynomial in x whose coefficients depend on r. Now, set this polynomial equal to zero and solve for r to find the characteristic equation.

Determining General Solution: Depending on the roots of the characteristic equation, you’ll have different forms for the general solution:

If the roots r₁ and r₂ are real and distinct, the general solution is y(x) = c_1x^{r_1} + c_2x^{r_2}.
If there is a repeated root r, the general solution is y(x) = c_1x^r + c_2x^r\ln(x).
If the roots are complex, say r = \alpha \pm \beta i, the general solution is y(x) = x^\alpha(c_1\cos(\beta \ln(x)) + c_2\sin(\beta \ln(x))).

Applying Initial/Boundary Conditions: If you have initial or boundary conditions, you can use them to solve for the constants c₁ and c₂.

Cauchy-Euler Equation Solved Problems

Question 1: Solve a Cauchy-Euler equation step by step. Consider the second-order Cauchy-Euler equation: x²y′′−6xy′+13y=0

Solution:

Assume a solution of the form y = x^r.
The derivatives are y’ = rx^{r-1} and y’’ = r(r-1)x^{r-2}.
Substitute y, y’ , and y’’ back into the original equation:
x^2(r(r−1)x^{r−2})−6x(rx^{r−1})+13x^r=0
This simplifies to:
r(r−1)x^r−6rx^r+13x^r=0
Factor out x^r since ( x \neq 0 ) and solve the characteristic equation:
r(r−1)−6r+13=0
⇒ r^2−7r+13=0
This is a quadratic equation in r .
Since the discriminant b^2 - 4ac is negative the roots of the characteristic equation are complex . Let’s find the roots:
r=\dfrac{7±\sqrt{49−4(1)(13)}}{2}
⇒ r=\dfrac{7±\sqrt{49−52}}{2}
⇒ r=\dfrac{7±\sqrt{-3}}{2}
⇒ r=\dfrac{7}2±\dfrac{\sqrt3i}2
The roots are complex hence, the general solution is:
y(x)=x^{7/2}(c_1cos(\dfrac{\sqrt3}2ln(x))+c_2sin(\dfrac{\sqrt3}2ln(x)))
Here, c_1 and c_2 are constants determined by boundary conditions or initial values.

Question 2: Solve the Cauchy-Euler equation step by step. Consider the second-order Cauchy-Euler equation: x²y’’ - 7xy’ + 16y = 0

Solution:

Let’s assume that y = x^ris the solution of the given differential equation, where (r) is a constant to be determined.
Substitute y = x^r into the differential equation:[ x^2y’’ - 7xy’ + 16y = 0] [x^2[r(r-1)x^{r-2}] - 7x[rx^{r-1}] + 16x^r = 0]
For the first derivative term: x^2[r(r-1)x^{r-2}] = r(r-1)x^r
For the second derivative term: 7x[rx^{r-1}] = 7rx^r
For the third term: 16x^r
Combining all terms: r(r-1)x^r - 7rx^r + 16x^r = 0
Set the polynomial equation equal to zero: r(r-1) - 7r + 16 = 0
Solving this quadratic equation for r: r^2 - 8r + 16 = 0
Factoring: (r-4)^2 = 0 The repeated root is r = 4.
Since we have a repeated root, the general solution is: y(x) = c_1x^4 + c_2x^4\ln(x)

Cauchy's Theorem
Euler Method for solving differential equation
Cauchy's Mean Value Theorem
Euler's Theorem

Practice Problems on Cauchy Euler Equation

Question 1: Solve the Cauchy-Euler differential equation: x²y′′+ 4xy′+ 4y = 0.

Question 2: Find the general solution of the Cauchy-Euler equation: x²y′′−3xy′+ y = 0.

Question 3: Determine the solution for the Cauchy-Euler differential equation: x²y′′+ 6xy′+ 9y = 0.

Question 4: Solve the Cauchy-Euler equation: x²y′′−2xy′+y = 0.

Question 5: Find the general solution of the Cauchy-Euler differential equation: x²y′′+5xy′+6y = 0.

Question 6: Solve the Cauchy-Euler differential equation: x²y′′+3xy′+y = 0.

Question 7: Determine the solution to the following Cauchy-Euler equation: x²y′′+ 2xy′−3y = 0.