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.