Step 7: Solve the Characteristic Equation
\( a m(m – 1) + b m + c = 0 \)
Comment: We rearrange terms to form a quadratic equation in terms of \( m \). This equation is known as the characteristic equation. Solving it will give us the roots that dictate the form of the general solution.

Step 8: Find Roots of Characteristic Equation
\( m = \frac{{-b \pm \sqrt{{b^2 – 4ac}}}}{2a} \)
Comment: Here, we apply the quadratic formula to find the roots of the characteristic equation. These roots can be real, complex, or repeated depending on the discriminant.

Step 9: Write Down the General Solution
\( y(x) = C_1 x^{m_1} + C_2 x^{m_2} \)

Comment: You might wonder why the general solution takes this form. It’s not arbitrary! The form is derived from our initial trial function \( y(x) = x^m \), which was chosen based on the form of Euler’s equation. Upon substituting this trial solution and its derivatives into the differential equation, we obtained a polynomial equation in terms of \( m \), called the characteristic equation. Solving this characteristic equation gives us roots \( m_1 \) and \( m_2 \). Thanks to the linearity of Euler’s equation, we know that if \( y_1 \) and \( y_2 \) are individual solutions, then a linear combination \( C_1 y_1 + C_2 y_2 \) will also be a solution. That’s why the general solution is \( y(x) = C_1 x^{m_1} + C_2 x^{m_2} \).

Step 1: Start with the Differential Equation
x²y” – 4xy’ + 6y = 0

Step 2: Assume a Trial Solution
Assume y(x) = xᵐ as a trial solution.

Step 3: Find the Derivatives
y'(x) = m xᵐ⁻¹ and y”(x) = m(m-1) xᵐ⁻²

Step 4: Substitute into the Equation
Substitute into our equation to get x²m(m-1)xᵐ⁻² – 4mxᵐ⁻¹ + 6xᵐ = 0.

Step 5: Simplify
Simplify to get m(m-1)xᵐ – 4mxᵐ + 6xᵐ = 0.

Step 6: Factor Out Common Terms
Factor out xᵐ to get (m(m-1) – 4m + 6)xᵐ = 0.

Step 7: Solve for Characteristic Equation
We get the characteristic equation m(m-1) – 4m + 6 = 0, simplifying to m² – m – 4m + 6 = 0 and finally m² – 5m + 6 = 0.

Step 8: Find Roots of Characteristic Equation
The roots are m₁ = 3 and m₂ = 2.

Step 9: Write Down the General Solution
y(x) = C₁ x³ + C₂ x²

Comment: The form of our general solution is derived directly from our initial assumptions and the roots of the characteristic equation. This is why y(x) = C₁ x³ + C₂ x² is the most general solution for the given Euler’s equation.

Step 1: Start with the Differential Equation
x²y” – 4xy’ + 4y = 0

Step 2: Assume a Trial Solution
Assume y(x) = xᵐ as a trial solution.

Step 3: Find the Derivatives
y'(x) = m xᵐ⁻¹, y”(x) = m(m-1) xᵐ⁻²

Step 4: Substitute into the Equation
x²m(m-1)xᵐ⁻² – 4mxᵐ⁻¹ + 4xᵐ = 0

Step 5: Simplify
m(m-1)xᵐ – 4mxᵐ + 4xᵐ = 0

Step 6: Factor Out Common Terms
(m(m-1) – 4m + 4)xᵐ = 0

Step 7: Solve for Characteristic Equation
m(m-1) – 4m + 4 = 0 ➞ m² – m – 4m + 4 = 0 ➞ m² – 5m + 4 = 0

Step 8: Find Roots of Characteristic Equation
m = 4 (repeated root)

Step 9: Write Down the General Solution
y(x) = C₁ x⁴ + C₂ x⁴ ln(x)

Comment: ln(x) appears in the solution due to the repeated root.

The Fundamental Choice of y(x) = xᵐ in Differential Equations

The choice of y(x) = xᵐ as the trial solution in solving differential equations is rooted in a combination of mathematical intuition and historical development.

Throughout the history of mathematics, mathematicians have sought methods to solve differential equations. When facing second-order linear homogeneous equations with constant coefficients, the trial solution y(x) = xᵐ was observed to provide a remarkably efficient and effective approach.

While there isn’t a more fundamental principle that dictates the choice of y(x) = xᵐ, the reasoning behind its effectiveness is based on empirical observation and mathematical convenience. Substituting y(x) = xᵐ often simplifies the equation and leads to solutions involving m.

This approach might not be derived from a single profound principle, but it’s a testament to the power of mathematical exploration and the ingenuity of mathematicians across different eras.

Historical Context:

The trial solution y(x) = xᵐ has been used since the early days of calculus and differential equations. It gained prominence as mathematicians sought systematic ways to solve differential equations with constant coefficients. This approach found its place alongside other techniques and tools in the mathematician’s toolkit, offering a reliable method for tackling a wide array of problems.