Consider the initial value problem y? + 16 y = e-t, y(0) = y0, y?(0) = . Suppose
ID: 3077794 • Letter: C
Question
Consider the initial value problem y? + 16 y = e-t, y(0) = y0, y?(0) = . Suppose we know that y (t) rightarrow 0 as t rightarrow infinity. Determine the solution and the initial conditions. y(t) = help (formulas) y(0) = help (numbers) y?(0) = help (numbers)Explanation / Answer
The general solution is y(t) = yh(t) + yp(t) where yp(t) is the particular solution and yh(t) is the homogeneous solution. In this case, the homogeneous solution is yh(t) = c1 cos(4t) + c2 sin(4t). If we know that y(t) -> 0 as t -> infinity, then it must be that c1 = c2 = 0!!! OR ELSE THAT PART WILL NEVER GO TO ZERO! OK, now we know that c1 = c2 = 0. The particular solution has form yp(t) = A e^(-t). Note that yp'' = A e^(-t) so that yp'' + 16 yp = 17 A e^(-t) Therefore (as written below) it must be that A = 1/17 and yp(t) = 1/17 e^(-t). The solution is (since yh(t) = 0) y(t) = 1/17 e^(-t) Note that y(0) = 1/17 and y'(0) = -1/17. Yes!