Final answer:
To solve the given differential equation y" - y' - 30y = 0 with the initial condition y(0) = 2, we assume a solution of the form y = e^(rt). Substituting this into the differential equation, factoring out e^(rt), and solving the resulting quadratic equation, we find the general solution y = c1e^(-5t) + c2e^(6t). Using the initial condition y(0) = 2, we can solve for the constants c1 and c2 to get the specific solution.
Step-by-step explanation:
The given initial value problem is y" - y' - 30y = 0, y(0) = 2.
To solve this differential equation, we can assume a solution of the form y = e^(rt).
Substituting this into the differential equation, we get r^2 e^(rt) - re^(rt) - 30e^(rt) = 0.
Factoring out e^(rt), we get e^(rt)(r^2 - r - 30) = 0.
Solving the quadratic equation r^2 - r - 30 = 0, we get r = -5 or r = 6.
Therefore, the general solution to the differential equation is y = c1e^(-5t) + c2e^(6t), where c1 and c2 are constants determined by the initial conditions.
Using the initial condition y(0) = 2, we can substitute t = 0 and y = 2 into the general solution to solve for the constants c1 and c2.
When t = 0, y = c1e^(0) + c2e^(0) = c1 + c2 = 2.
Since the equation c1 + c2 = 2 has infinitely many solutions, we need an additional condition to uniquely determine the values of c1 and c2.