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.