Question

In each of Problems 11 through 15 , find the solution of the given initial value problem.y+2y+5y=4e−tcos(2t),y(0)=1,y (0)=0

Answer 1

The solution to the given initial value problem
`y + 2y' + 5y = 4e^(-t)cos(2t), y(0) = 1, y'(0) = 0 is: y(t) = e^(-t)(cos(2t) + sin(2t)).`

Step-by-step explanation:

To solve this linear differential equation with initial values, we'll first find the complementary function (CF) and particular integral (PI). The characteristic equation for the homogeneous part is
`λ^2 + 2λ + 5 = 0,` which yields complex roots (λ = -1 + 2i) and (λ = -1 - 2i). Hence, the CF is
`y_c(t) = e^(-t)(Acos(2t) + Bsin(2t))`.

For the particular solution, assume
`y_p(t) = Ce^(-t)cos(2t) + De^(-t)sin(2t)`. By differentiating and substituting into the differential equation, we find
`y_p(t) = e^(-t)(cos(2t) + sin(2t))`. Thus, the general solution is
`y(t) = y_c(t) + y_p(t)`.

Applying the initial conditions y(0) = 1 and y'(0) = 0 to the general solution, we solve for the constants A and B from the CF and C and D from the PI. Since
`y'(t) = -e^(-t)(cos(2t) - sin(2t))`, we substitute these into the initial conditions, leading to A = 1 and C = 1.

Therefore, the final solution with the given initial values is
`y(t) = e^(-t)(cos(2t) + sin(2t))`, fulfilling the initial value problem conditions.

This solution represents the function that satisfies the given differential equation and the provided initial conditions, describing the behavior of y(t) over time in response to the differential equation's dynamics.

Answer 2

The question pertains to college-level Physics, focusing on wave functions and the linear wave equation, and involves solving for the superposition of two wave functions with a phase shift.

Step-by-step explanation:

The subject of this question is Physics, specifically, it deals with the concepts of wave functions and the linear wave equation.

The questions provided seem to come from a college-level physics textbook that covers wave dynamics.

In one of the problems mentioned, given two wave functions that differ by a phase shift, where one is

y1 (x, t) = A cos (kx - wt)

and the other is y2 (x, t) = A cos (kx - wt + p),

one can find the resulting wave from the superposition of these two waves using trigonometric identities.

The formula cos u + cos v = 2 cos (½(u + v)) cos (½(u - v)) can be used to express the superposition in terms of a single wave function, which simplifies the analysis.