Final answer:
To solve the given initial value problem y'' + 10y' + 25y = 0, we can assume a solution of the form y(t) = e^(rt), where r is a constant. The general solution for y(t) is y(t) = c1e^(-5t) + c2te^(-5t). Using the initial conditions y(0) = 2 and y'(0) = -7, we can find the specific values of c1 and c2.
Step-by-step explanation:
To solve the given initial value problem y'' + 10y' + 25y = 0, we can assume a solution of the form y(t) = e^(rt), where r is a constant. By substituting this solution into the differential equation, we obtain the characteristic equation r^2 + 10r + 25 = 0. Solving this quadratic equation gives us r = -5. Therefore, the general solution for y(t) is y(t) = c1e^(-5t) + c2te^(-5t).
Using the initial conditions y(0) = 2 and y'(0) = -7, we can find the specific values of c1 and c2. Substituting t = 0 and y = 2 into the general solution, we get 2 = c1e^(0) + c2(0)e^(0), which simplifies to c1 = 2. Substituting t = 0 and y' = -7 into the general solution, we get -7 = -5c1e^(0) + c2(1)e^(0), which simplifies to c2 = -5.
Therefore, the solution to the initial value problem y'' + 10y' + 25y = 0, y(0) = 2, y'(0) = -7 is y(t) = 2e^(-5t) - 5te^(-5t).