Answer Happy

where t0 is a positive constant
Need help with part (c)
x' = ax(t – b), (0.2) where a, b are constants, has a solution of the form y(t) = Cest for any constant C, provided s satisfies the transcendental equation s = ae-bs. A solution to (0.2) for t > 0 can also be found using the method of steps. Assume that r(t) = f(t) for –b <t<0. For 0 <t<b, equation (0.2) becomes x(t) = axít – b) = afệt – b), = and so X(t) = - jasw- af (v – b)dv + z(0) v - b. Now that we know y(t) on (0,b], we can repeat this procedure to obtain a(t) = | ax(v – bydv + 2(1) (xb b for b < x < 26. This process can be continued indefinitely.
(b) Use the method of steps to show that the solution to the initial value problem '(t) = -2(t-1), = x(t) = 1 on (-1,0), is given by n x(t) = Ż(-1)« [t – (k – 1))" = 3- for n-1<t<n, k! k=0 where n is a nonnegative integer. (c) Use the method of steps to compute the solution to the initial value problem given in (0.1) on the interval 0 <t < 15 for to = 3. = (0.1) 3 x' (t) = 6 - x (t – to), 500 x(t) = 0 for x E[-to, 0], =