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The given differential equation is:

\[
25y'' - 4y = 0
\]

This is a second-order linear homogeneous differential equation with constant coefficients. Let's solve it step-by-step.

### Step 1: Form the characteristic equation
Assume the solution is of the form \( y(t) = e^{rt} \). Substituting \( y = e^{rt} \), \( y' = re^{rt} \), and \( y'' = r^2 e^{rt} \) into the differential equation:

\[
25r^2 e^{rt} - 4e^{rt} = 0
\]

Dividing both sides by \( e^{rt} \) (since \( e^{rt} \neq 0 \)):

\[
25r^2 - 4 = 0
\]

### Step 2: Solve the characteristic equation
\[
25r^2 = 4
\]
\[
r^2 = \frac{4}{25}
\]
\[
r = \pm \frac{2}{5}
\]

### Step 3: Write the general solution
Since the characteristic equation has two distinct real roots \( r_1 = \frac{2}{5} \) and \( r_2 = -\frac{2}{5} \), the general solution to the differential equation is:

\[
y(t) = c_1 e^{\frac{2}{5}t} + c_2 e^{-\frac{2}{5}t}
\]

where \( c_1 \) and \( c_2 \) are arbitrary constants.

### Final Answer:
\[
y(t) = c_1 e^{\frac{2}{5}t} + c_2 e^{-\frac{2}{5}t}
\]

Would you like more details or have any questions about this solution?

Here are some related questions:
1. How do you solve homogeneous second-order differential equations with complex roots?
2. What happens if the characteristic equation has a double root?
3. How would the solution change if this were a non-homogeneous differential equation?
4. How do initial conditions affect the solution of second-order differential equations?
5. How can you use Laplace transforms to solve differential equations?

**Tip:** For differential equations with constant coefficients, always start by forming the characteristic equation.

Find the general solution of the following differential equation: 25y'' - 4y = 0

Learn how to solve the second-order linear homogeneous differential equation 25y'' - 4y = 0 using characteristic equations and exponential functions. This step-by-step guide explains the use of constants and is suitable for students studying undergraduate-level calculus.

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