Explanation:
The given differential equation is a homogeneous linear second-order differential equation with constant coefficients. We can find the general solution by assuming a solution of the form y = e^(rt), where r is a constant.
Substituting this into the differential equation, we get:
r^2 e^(rt) - 6r e^(rt) + 90 e^(rt) = 0
Dividing both sides by e^(rt), we get:
r^2 - 6r + 90 = 0
Solving for r using the quadratic formula, we get:
r = (6 ± sqrt(6^2 - 4*90)) / 2
r = 3 ± 3i√7
Since the roots are complex, the general solution takes the form:
y(t) = e^(3t)(c1 cos(3√7 t) + c2 sin(3√7 t))
To find the particular solution that satisfies the initial conditions, we need to solve for the constants c1 and c2. Taking the first derivative of y(t), we get:
y'(t) = 3e^(3t)(c1 cos(3√7 t) + c2 sin(3√7 t)) + e^(3t)(-3√7 c1 sin(3√7 t) + 3√7 c2 cos(3√7 t))
Evaluating y'(0) using the initial condition, we get:
y'(0) = 3c1 + 10√7 c2 = 10
Taking the second derivative of y(t), we get:
y''(t) = 9e^(3t)(c1 cos(3√7 t) + c2 sin(3√7 t)) + 6e^(3t)(-√7 c1 sin(3√7 t) + √7 c2 cos(3√7 t)) - 21e^(3t)(c1 cos(3√7 t) + c2 sin(3√7 t))
Evaluating y''(0) using the initial condition, we get:
y''(0) = 9c1 - 21c1 = -12c1 = -72
Therefore, c1 = 6 and c2 = (10 - 3c1)/(√7) = (10 - 18)/(√7) = -8/√7.
The particular solution that satisfies the initial conditions is:
y(t) = e^(3t)(6 cos(3√7 t) - 8/√7 sin(3√7 t))
Therefore, the solution to the initial value problem is:
y(t) = e^(3t)(6 cos(3√7 t) - 8/√7 sin(3√7 t)), y(0) = 8, y'(0) = 10.