How do I find the general solution of the second order differential equation y” = 25y?

To find the general solution of the second order differential equation given by y” = 25y, we start by rewriting it in a more standard form:

y” – 25y = 0

This is a linear homogeneous differential equation with constant coefficients. We can solve it using the characteristic equation method. The characteristic equation corresponding to this differential equation is:

r² – 25 = 0

Now, we solve for r:
r² = 25
r = ±5

Thus, we have two distinct real roots: r₁ = 5 and r₂ = -5.

The general solution for a second order linear differential equation with distinct roots can be expressed as:

y(t) = C₁e^r₁t + C₂e^r₂t

Substituting the values of the roots, we get:

y(t) = C₁e^5t + C₂e^-5t

Where C₁ and C₂ are arbitrary constants determined by the initial or boundary conditions of the problem.

This general solution describes the behavior of the system represented by the differential equation over time.