# Solve the second-order ODE: \(y” – 4y = 0\)

**1. Write down the given differential equation:**

\[

y” – 4y = 0

\]

**2. Identify the characteristic equation:**

The characteristic equation associated with this second-order ODE is obtained by substituting \(y = e^{rx}\) into the equation:

\[

r^2 e^{rx} – 4 e^{rx} = 0

\]

**3. Simplify the characteristic equation:**

Factor out \(e^{rx}\):

\[

e^{rx} (r^2 – 4) = 0

\]

**4. Solve the characteristic equation:**

Set the equation equal to zero:

\[

r^2 – 4 = 0

\]

Solve for \(r\):

\[

r^2 = 4

\]

\[

r = \pm 2

\]

**5. Write down the solutions for \(r\):**

The solutions to the characteristic equation give us the roots \(r_1 = 2\) and \(r_2 = -2\).

**6. Write down the general solution:**

Since the roots are distinct real numbers, the general solution to the ODE is:

\[

y(x) = C_1 e^{2x} + C_2 e^{-2x}

\]

where \(C_1\) and \(C_2\) are arbitrary constants to be determined by initial conditions.

**Final Answer**

The general solution to the differential equation \(y” – 4y = 0\) is:

### \[

y(x) = C_1 e^{2x} + C_2 e^{-2x}

\]

where \(C_1\) and \(C_2\) are arbitrary constants.

This solution represents a linear combination of exponential functions, reflecting the two independent solutions derived from the characteristic equation.