How to Solve Homogeneous Cauchy-Euler Equations: A Step-by-Step Guide

📚 Understanding Homogeneous Cauchy-Euler Equations

A Homogeneous Cauchy-Euler equation, also known as an Euler equation, is a type of linear differential equation with variable coefficients. It has a specific form that allows for a standard solution method. These equations frequently appear in physics and engineering.

📜 A Brief History

The Cauchy-Euler equation is named after Augustin-Louis Cauchy and Leonhard Euler, who were pioneers in the study of differential equations. Euler's work in the 18th century laid the foundation for solving these types of equations, while Cauchy further developed the theory in the 19th century.

🔑 Key Principles

• 🔍 General Form: A homogeneous Cauchy-Euler equation has the form $ax^2\frac{d^2y}{dx^2} + bx\frac{dy}{dx} + cy = 0$, where $a$, $b$, and $c$ are constants.
• 💡 Substitution: The standard method to solve these equations involves the substitution $x = e^t$, which transforms the equation into a linear differential equation with constant coefficients.
• 📝 Derivatives: We need to express the derivatives $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$ in terms of $\frac{dy}{dt}$ and $\frac{d^2y}{dt^2}$. Using the chain rule:
  • $\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx} = \frac{1}{x} \frac{dy}{dt} = e^{-t} \frac{dy}{dt}$
  • $\frac{d^2y}{dx^2} = \frac{d}{dx}(\frac{dy}{dx}) = \frac{d}{dx}(e^{-t} \frac{dy}{dt}) = e^{-2t}(\frac{d^2y}{dt^2} - \frac{dy}{dt})$
• 🔢 Characteristic Equation: After substitution, the transformed differential equation can be solved by finding the roots of its characteristic equation.
• 🌱 General Solution: The general solution depends on the nature of the roots of the characteristic equation (real and distinct, repeated real roots, or complex conjugate roots).

✍️ Step-by-Step Solution

1. Given Equation: $ax^2\frac{d^2y}{dx^2} + bx\frac{dy}{dx} + cy = 0$
2. Substitution: Let $x = e^t$, so $t = \ln(x)$.
3. Compute Derivatives:
   • $\frac{dy}{dx} = e^{-t} \frac{dy}{dt}$
   • $\frac{d^2y}{dx^2} = e^{-2t}(\frac{d^2y}{dt^2} - \frac{dy}{dt})$
4. Substitute into the Equation: $a(e^t)^2 e^{-2t}(\frac{d^2y}{dt^2} - \frac{dy}{dt}) + b(e^t)e^{-t}\frac{dy}{dt} + cy = 0$ Simplifies to: $a(\frac{d^2y}{dt^2} - \frac{dy}{dt}) + b\frac{dy}{dt} + cy = 0$ $a\frac{d^2y}{dt^2} + (b-a)\frac{dy}{dt} + cy = 0$
5. Solve the Transformed Equation:
   • Form the characteristic equation: $ar^2 + (b-a)r + c = 0$
   • Solve for $r$ using the quadratic formula.
6. General Solution Based on Roots:
   • Distinct Real Roots ($r_1 \neq r_2$): $y(t) = c_1e^{r_1t} + c_2e^{r_2t}$. Substitute back $t = \ln(x)$ to get $y(x) = c_1x^{r_1} + c_2x^{r_2}$.
   • Repeated Real Roots ($r_1 = r_2 = r$): $y(t) = (c_1 + c_2t)e^{rt}$. Substitute back $t = \ln(x)$ to get $y(x) = (c_1 + c_2\ln(x))x^r$.
   • Complex Conjugate Roots ($r = \alpha \pm i\beta$): $y(t) = e^{\alpha t}(c_1\cos(\beta t) + c_2\sin(\beta t))$. Substitute back $t = \ln(x)$ to get $y(x) = x^{\alpha}(c_1\cos(\beta \ln(x)) + c_2\sin(\beta \ln(x)))$.

🧮 Example

Solve $x^2y'' - 2xy' + 2y = 0$.

1. Equation: $x^2y'' - 2xy' + 2y = 0$
2. Characteristic Equation: $r(r-1) - 2r + 2 = 0 \Rightarrow r^2 - 3r + 2 = 0$
3. Solve for r: $(r-1)(r-2) = 0 \Rightarrow r_1 = 1, r_2 = 2$
4. General Solution: $y(x) = c_1x + c_2x^2$

🧪 Real-World Applications

• ⚙️ Engineering Mechanics: Analyzing stress distribution in structures.
• ⚡ Electrical Circuits: Modeling circuits with variable inductances.
• 🌊 Fluid Dynamics: Describing certain fluid flow problems.

✅ Conclusion

Homogeneous Cauchy-Euler equations are a fascinating and practical area of differential equations. By understanding the substitution method and the nature of the characteristic roots, you can effectively solve these equations and apply them to various real-world problems.