Solving (x^2D^2 - 4xD + 6)y = X^2 A Cauchy-Euler Method Guide

In the realm of differential equations, the Cauchy-Euler equation, also known as the Euler-Cauchy equation, stands as a prominent type of linear differential equation with variable coefficients. These equations exhibit a specific form where the coefficients are polynomials in the independent variable. This article delves into the intricacies of solving the differential equation extbf{((x^2D2 - 4xD + 6)y = x^2)} using the powerful Cauchy-Euler method. This method provides a systematic approach to finding solutions for this class of equations, which frequently arise in various fields of science and engineering. The Cauchy-Euler equation is a special type of linear differential equation with variable coefficients that can be transformed into a linear differential equation with constant coefficients by a suitable change of variable.

Understanding Cauchy-Euler Equations

To effectively tackle the given differential equation, it's crucial to grasp the essence of Cauchy-Euler equations. These equations possess a distinctive form:

ax^2 \frac{d^2y}{dx2} + bx \frac{dy}{dx} + cy = f(x)

where a, b, and c are constants. The key characteristic of these equations lies in the fact that the power of x in each term matches the order of the derivative. This property allows us to employ a clever substitution that transforms the equation into a more manageable form with constant coefficients. The Cauchy-Euler equation plays a pivotal role in various applications, including modeling physical systems with varying properties and analyzing the behavior of electrical circuits. This equation's unique structure allows for a systematic solution approach, making it a valuable tool for engineers and scientists.

Method to Solve

The standard method to solve Cauchy-Euler equations involves assuming a solution of the form y = x^m, where m is a constant to be determined. By substituting this assumed solution and its derivatives into the original equation, we obtain an auxiliary equation in terms of m. The roots of this auxiliary equation dictate the form of the general solution. If the roots are distinct and real, the general solution is a linear combination of terms of the form x^m. If the roots are complex conjugates, the general solution involves trigonometric functions. And if the roots are repeated, the general solution includes logarithmic terms.

Transforming the Equation

The equation we aim to solve is extbf{((x^2D2 - 4xD + 6)y = x^2)}. Here, D represents the differential operator d/dx. To align this equation with the standard Cauchy-Euler form, we introduce a substitution: x = e^t. This substitution is the cornerstone of the Cauchy-Euler method, as it elegantly transforms the variable coefficient equation into a constant coefficient one.

Substitution

With this substitution, we can express the derivatives with respect to x in terms of derivatives with respect to t. Let's denote dy/dt as y' and d^2y/dt2 as y''. Using the chain rule, we have:

$\frac{dy}{dx} = \frac{dy}{dt} \frac{dt}{dx} = \frac{1}{x} \frac{dy}{dt} = e^{-t}y'$

$\frac{d^2y}{dx^2} = \frac{d}{dx}(\frac{dy}{dx}) = \frac{d}{dx}(e^{-t}y') = \frac{d}{dt}(e^{-t}y')\frac{dt}{dx} = e^{-2t}(y'' - y')$

Transformed Equation

Substituting these expressions into the original equation, we get:

e^{2t}(e{-2t}(y'' - y')) - 4e^t(e{-t}y') + 6y = e^{2t}

Simplifying, we obtain a linear differential equation with constant coefficients:

y'' - 5y' + 6y = e^{2t}

This transformed equation is significantly easier to solve than the original Cauchy-Euler equation. The substitution x = e^t has effectively converted the variable coefficient problem into a constant coefficient one, which we can solve using standard techniques.

Solving the Transformed Equation

Now, we focus on solving the transformed equation:

y'' - 5y' + 6y = e^{2t}

This is a second-order linear non-homogeneous differential equation with constant coefficients. To solve it, we first find the homogeneous solution and then a particular solution. The general solution is the sum of the homogeneous and particular solutions.

Homogeneous Solution

To find the homogeneous solution, we consider the homogeneous equation:

y'' - 5y' + 6y = 0

We assume a solution of the form y = e^{rt}, where r is a constant. Substituting this into the homogeneous equation, we get the characteristic equation:

r^2 - 5r + 6 = 0

Factoring the quadratic, we find the roots:

(r - 2)(r - 3) = 0

r = 2, 3

Since we have two distinct real roots, the homogeneous solution is:

y_h = c_1e^{2t} + c_2e^{3t}

where c_1 and c_2 are arbitrary constants. The homogeneous solution represents the natural response of the system without the external forcing function.

Particular Solution

Next, we find a particular solution to the non-homogeneous equation:

y'' - 5y' + 6y = e^{2t}

Since the forcing function is e^{2t}, we assume a particular solution of the form y_p = Ate^{2t}, where A is a constant to be determined. The inclusion of the factor t is necessary because e^{2t} is already part of the homogeneous solution. Taking the first and second derivatives of y_p, we get:

y_p' = A(e^{2t} + 2te^{2t})

y_p'' = A(4e^{2t} + 4te^{2t})

Substituting these into the non-homogeneous equation, we have:

A(4e^{2t} + 4te^{2t}) - 5A(e^{2t} + 2te^{2t}) + 6Ate^{2t} = e^{2t}

Simplifying, we get:

-Ae^{2t} = e^{2t}

Thus, A = -1. Therefore, the particular solution is:

y_p = -te^{2t}

The particular solution represents the system's response to the specific external forcing function.

General Solution

The general solution is the sum of the homogeneous and particular solutions:

y(t) = y_h + y_p = c_1e^{2t} + c_2e^{3t} - te^{2t}

This general solution represents the complete response of the system, including both the natural response and the response to the external forcing function. The arbitrary constants c_1 and c_2 can be determined by applying initial conditions, if provided.

Transforming Back to the Original Variable

Finally, we transform the solution back to the original variable x using the substitution t = ln(x):

y(x) = c_1e^{2ln(x)} + c_2e^{3ln(x)} - ln(x)e^{2ln(x)}

Using the property e^{ln(x)} = x, we simplify this to:

y(x) = c_1x^2 + c_2x^3 - x^2ln(x)

This is the general solution to the original Cauchy-Euler equation. It represents the family of functions that satisfy the given differential equation. The constants c_1 and c_2 can be determined if initial conditions or boundary conditions are provided.

Verifying the Solution

To ensure the correctness of the solution, it's always a good practice to verify it by substituting it back into the original differential equation. This involves computing the first and second derivatives of the solution and plugging them into the equation. If the equation holds true, then the solution is verified.

Conclusion

In conclusion, we have successfully solved the differential equation extbf{((x^2D2 - 4xD + 6)y = x^2)} using the Cauchy-Euler method. This method involves a clever substitution that transforms the equation into a constant coefficient one, which can be solved using standard techniques. The solution obtained is:

y(x) = c_1x^2 + c_2x^3 - x^2ln(x)

where c_1 and c_2 are arbitrary constants. The Cauchy-Euler equation is a powerful tool for solving a specific class of differential equations with variable coefficients. Its applications span various fields, making it an essential concept in mathematics, physics, and engineering. By mastering the Cauchy-Euler method, one can effectively tackle a wide range of problems involving differential equations with variable coefficients.

The Cauchy-Euler method provides a systematic approach to solving these equations, making them more accessible and manageable. By understanding the underlying principles and applying the steps outlined in this article, one can confidently tackle Cauchy-Euler equations and similar problems in various scientific and engineering contexts. This article serves as a comprehensive guide to understanding and applying the Cauchy-Euler method for solving differential equations. The systematic approach, the detailed explanations, and the step-by-step solution make it a valuable resource for students, engineers, and anyone interested in mastering this important mathematical technique.