set 15 problem 3 Consider the following initial value proble

set 15 problem 3

Consider the following initial value problem: y\" + 49y = {5, 0 lessthanorequalto t lessthanorequalto 3 0, t > 3 y(0) = 5, y\'(0) 0 Using Y for the Laplace transform of y(t), i.e., Y = L{y(t)}, find the equation you get by taking the Laplace transform of the differential equation and solve for Y(s) =

Solution

we have y \'\' + 49 y = g(t)
where g(t) = 5 for 0<=t<=3, g(t)=0 for t>3,

and y(0) =5 , y\'(0) = 0

call Y(s) = L(y(t)).
we have: L(49y) = 0 L(y) = 49 Y(s)
and: L(y \'\') = s^2 Y(s) - s y(0) - y\'(0) = s^2 Y(s)
so: L(y \'\' + 49y) = (s^2 + 49) Y(s)

Now: Integral{}
G(s) = L(g(t)) = Integral{t=0 to inf; g(t) exp(-st) dt}
= Integral{t=0 to 3; g(t) exp(-st) dt} + Integral{t=3 to inf; g(t) exp(-st) dt}
= Integral{t=0 to 3; 5 exp(-st) dt} + Integral{t=3 to inf; 0 exp(-st) dt}
= 5t\' Integral{t=0 to 3; t exp(-st) dt} + 0 Integral{t=3 to inf; exp(-st) dt}
I trust you can find the primitives of t exp(-st) and exp(-st), which are simple, and we obtain:
G(s) = 5t\' { 1/s^2 - (3/s) exp(-3s) - (1/s^2) exp(-3s) } + 0 { (1/s) exp(-3s) }
= 5t\' (1 - exp(-3s) ) / s^2

So the LT of the differential equation is:

(s^2 + 49) Y(s) = 5t\' (1 - exp(-3s) ) / s^2
and isolating Y(s) gives us:

Y(s) = 5t\' [1 - exp(-3s)] / [s^2 (s^2 + 49)] = (5t\'/49) [1 - exp(-3s)] [1/s^2 - 1/(s^2 + 49) ]