Final answer:
To solve the second-order linear differential equation y'' + 4y = 0 with initial conditions y(0) = 7 and y'(0) = -10, we use the characteristic equation method. We find the particular solution is y(t) = 7cos(2t) - 5sin(2t) by applying the initial conditions to determine the constants in the general solution.
Step-by-step explanation:
The second-order linear differential equation you're looking to solve is y'' + 4y = 0 with the initial conditions y(0) = 7 and y'(0) = -10. We'll solve this using the characteristic equation method.
First, we find the characteristic equation by substituting y = ert, where r is a constant, which gives us r2 + 4 = 0. This has roots r1 = 2i and r2 = -2i, indicating a solution of the form y(t) = C1cos(2t) + C2sin(2t), where C1 and C2 are constants to be determined.
Using the initial condition y(0) = 7, we find that C1 = 7. To find C2, differentiate y(t) to obtain y'(t) = -2C1sin(2t) + 2C2cos(2t), and then use the initial condition y'(0) = -10, resulting in C2 = -5. The final solution to the differential equation is y(t) = 7cos(2t) - 5sin(2t).