Consider the initial value problem where g(t) = y″ + 36y = g(t), if 0 < t < 5 \0_if 5≤t<∞. a. Take the Laplace transform of both sides of the given differential equation to create the corresponding algebraic equation. Denote the Laplace transform of y(t) by Y (s). Do not move any terms from one side of the equation to the other (until you get to part (b) below). (s^2+36)Y(s) 1/s^2 y(0) = 0, y'(0) = 0, b. Solve your equation for Y(s). Y(s) = L{y(t)} = = 1/36[1/s^2-1/(s^2+36)] = help (formulas) c. Take the inverse Laplace transform of both sides of the previous equation to solve for y(t). y(t) = 1/36[t-sin(6t)/6]

dv A model for the velocity v at timetof a certain object faling under the influence cf gravity in a viscous medium is given by the equation 1-. From the direction field shown in the figure to the right, skotch the dt 12 solutions with the initial conditions v(0) = 10, 12, and 19. Why is the value v= 12 calied the "terminal velocity"? Choose the correct skelch of the solutions with the initial conditions v(0) = 10, 12, and 19. O A. OB. Oc.

Two numbers are greater than 10 but less than 40. Their GCF is 17. What are the two numbers?

Solve the differential equation using either the method of undetermined coefficients or Laplace transforms. y'''+y"+y'+y=e-t+4t