Question

Solve for Y(s), the Laplace transform of the solution y(t) to the initial value problem below. y′′−4y′+4y=cos5t−sin5t,y(0)=4,y′(0)=2

Y(s)=

Answer 1

To find Y(s), the Laplace transform of the solution y(t) to the initial value problem, we take the Laplace transform of the differential equation and solve for Y(s). The Laplace transform of y'' - 4y' + 4y is s^2Y(s) - sy(0) - y'(0) - 4sY(s) + 4Y(s), and the Laplace transform of the right-hand side is 1/(s^2 + 25). By setting these equal to each other and rearranging, we can solve for Y(s).

Step-by-step explanation:

To solve for Y(s), the Laplace transform of the solution y(t) to the initial value problem, we first need to take the Laplace transform of the differential equation. Using the linearity property of the Laplace transform, we find that the Laplace transform of y'' - 4y' + 4y is s^2Y(s) - sy(0) - y'(0) - 4sY(s) + 4Y(s). Taking the Laplace transform of the right-hand side, which is cos(5t) - sin(5t), gives us 1/(s^2 + 25). Setting the two sides equal to each other, we can solve for Y(s).

Combining like terms and rearranging, we get (s^2 - 4s + 4)Y(s) = s + 4 - 1/(s^2 + 25). Factoring the left-hand side, we have (s - 2)^2Y(s) = s + 4 - 1/(s^2 + 25). Finally, solving for Y(s), we get Y(s) = (s + 4 - 1/(s^2 + 25))/(s - 2)^2. This is the Laplace transform of the solution y(t) to the initial value problem.