Final answer:
The initial value problem for the differential equation y'' - 5y' + 4y = 0 with initial conditions y(0) = -1 and y'(0) = -5 is solved by finding the characteristic equation, its roots, and applying initial conditions to determine the constants. The solution is y(t) = -e^{4t}.
Step-by-step explanation:
To solve the differential equation y'' - 5y' + 4y = 0 with initial conditions y(0) = -1 and y'(0) = -5, we begin by finding the characteristic equation of the homogeneous differential equation, which is given by r^2 - 5r + 4 = 0. Factoring this quadratic, we get (r - 4)(r - 1) = 0. Hence, the two roots are r = 4 and r = 1.
These roots suggest that the general solution to the differential equation is y(t) = C1e^{4t} + C2e^{t}, where C1 and C2 are constants to be determined using the initial conditions.
To find C1 and C2, we use the initial conditions:
- y(0) = -1 Gives us C1 + C2 = -1
- Computing y'(t) = 4C1e^{4t} + C2e^{t} and using y'(0) = -5 gives us 4C1 + C2 = -5
Solving the system of equations for C1 and C2, we find C1 = -1 and C2 = 0. Therefore, the particular solution with the initial conditions is y(t) = -e^{4t}.