The given differential equation is a second order homogeneous differential equation in the standard form ay'' + by = 0, where a = 25, b = -9 and y'' is the second derivative of y with respect to t.
This type of differential equation has a general solution in the form: y(t) = C1*cos(sqrt(b/a)*t) + C2*sin(sqrt(b/a)*t)
So, in our case, a = 25 and b = -9, so the general solution of the differential equation is:
y(t) = C1*cos(sqrt(-9/25)*t) + C2*sin(sqrt(-9/25)*t)
Now, we have to solve for C1 and C2 using the initial (boundary) conditions y(0) = 4 and y'(0) = 8.
For y(0) = 4:
y(t) = C1*cos(0) + C2*sin(0) = 4
This means that C1 = 4
For y'(0) = 8, we first need to find the derivative of y(t) with respect to t:
y'(t) = -C1*sqrt(-9/25)*sin(sqrt(-9/25)*t) + C2*sqrt(-9/25)*cos(sqrt(-9/25)*t)
When we substitute t=0, y'(0) = 8
we get:
8 = -C1*0+ C2*sqrt(-9/25)*1
So, C2 = 40/3
Therefore, the solution to the given differential equation with initial conditions is:
y(t) = 4cos(sqrt(-9/25)*t) + 40/3*sin(sqrt(-9/25)*t)
This simplifies to:
y(t) = 4cos((3/5)t) + (40/3)sin((3/5)t)