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The given differential equation is:

\[
\frac{d^2 y}{dx^2} - 5 \frac{dy}{dx} + 6y = 0
\]

This is a second-order linear homogeneous differential equation with constant coefficients. The general method to solve such equations involves finding the characteristic equation, solving for its roots, and using these roots to construct the general solution.

### Step-by-step solution:

1. **Characteristic Equation**:
   Replace \(\frac{d^2 y}{dx^2}\) with \(r^2\), \(\frac{dy}{dx}\) with \(r\), and \(y\) with 1 to form the characteristic equation:
   \[
   r^2 - 5r + 6 = 0
   \]

2. **Solve the Characteristic Equation**:
   Factor the quadratic equation:
   \[
   (r - 2)(r - 3) = 0
   \]
   Thus, the roots are \(r = 2\) and \(r = 3\).

3. **General Solution**:
   Since both roots are real and distinct, the general solution to the differential equation is:
   \[
   y(x) = C_1 e^{2x} + C_2 e^{3x}
   \]
   where \(C_1\) and \(C_2\) are arbitrary constants determined by initial conditions (if given).

Would you like a more detailed explanation, or do you have any specific questions about the process?

Here are 5 related questions for further practice:
1. What happens if the characteristic equation has repeated roots?
2. How would you solve a non-homogeneous version of this differential equation?
3. What are the solutions if the characteristic equation has complex roots?
4. How can initial conditions be used to determine the constants \(C_1\) and \(C_2\)?
5. How does changing the coefficients in the differential equation affect the general solution?

**Tip**: For second-order differential equations, always start by finding the characteristic equation, as it simplifies the problem considerably.

Solve the differential equation: d^2y/dx^2 - 5 dy/dx + 6y = 0.

This page provides a step-by-step guide to solving the second-order linear homogeneous differential equation with constant coefficients: d^2y/dx^2 - 5 dy/dx + 6y = 0. Learn how to find the characteristic equation and solve for real distinct roots. Suitable for undergraduate students.

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