Respuesta :
In order to solve this IVP using Laplace transforms, we must first write f(t) in terms of the Heaviside function.
f(t)=0*(u(t)-u(t-Pi))+1*(u(t-Pi)-u(t-2Pi))+0*(u(t-2Pi))
f(t)=u(t-π)-u(t-2π)
So, the rewritten IVP is
y'' +y = u(t-π)-u(t-2π)y(0)=0, y'(0)=1
Taking the Laplace transform of both sides of the equation, we get:
s2L{y}-sy(0)-y'(0)+L{y}=(1/s)*e-πs-(1/s)*e-2πs
s2L{y}-1+L{y}=(1/s)*e-πs-(1/s)*e-2πs
(s2+1)L{y}=1+(1/s)*e-πs-(1/s)*e-2πs
L{y}=1/(s2+1)+(1/s(s2+1))e-πs-(1/s(s2+1))*e-2πs
Now, we must take the inverse transform of both sides to solve for y.
The first inverse transform is easy enough. By definition, it is sin(t).
The second two inverse transforms will be a little tougher, we will have to use partial fraction decomposition to break them down into terms that are easier to compute.
A/s+(Bs+C)/(s^2+1)=1/(s(s^2+1))
A(s^2+1)+(Bs+C)(s)=1
As^2+A+Bs^2+Cs=1
Rewriting this system in matrix form, we get:
1 1 0 A 0
0 0 1 * B = 0
1 0 0 C 1
Using row-reduction we find that A=1, B=-1, and C=0. So, our reduced inverse transforms are:
L-1{(e-πs)(1/s-s/(s2+1))}
and
L-1{(e-2πs)(1/s-s/(s2+1))}
Using the first and second shifting properties, these inverse transforms can be computed as.
L-1{(e-πs)(1/s-s/(s2+1))}=u(t-π)-cos(t-π)u(t-π)
L-1{(e-2πs)(1/s-s/(s2+1))}=u(t-2π)-cos(t-2π)u(t-2π)
Combining all of our inverses transforms, we get the solution the IVP as:
y=sin(t)+u(t-π)-cos(t-π)u(t-π)+u(t-2π)-cos(t-2π)u(t-2π)
In mathematics, the Laplace transform, named after its discoverer Pierre Simon Laplace (/ləˈplɑːs/), transforms a function of real variables (usually in the time domain) into a function of complex variables (in the time domain). is the integral transform that Complex frequency domain, also called S-area or S-plane).
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