- #1
wahoo2000
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Homework Statement
Solve the IVP for t>0
[tex]
y'(t)+\int^{t}_{0}y(\tau)d\tau =
[/tex]
=sin(t) for 0<t<=pi
=0 for t>pi
y(0)=-1
The Attempt at a Solution
The solution depends on how large t is and therefore the solution consist of two parts depending on the size of t..
Let's call them y1(t) (for 0<t<=pi) and y2(t) (for t>pi)
This is solved using Laplace transform.
for y2, after Laplacetransform, Y(s) is called just Y for convenience.
sY-y(0)+Y/s=0
[tex]Y=\frac{-1}{s+1/s}=...simplification...=\frac{-s}{s^{2}+1}[/tex]
Then using inverse lapcace transform
==> y2(t)=-cos(t)
then for y1:
[tex]Y=\frac{\frac{1}{s^{2}+1}-1}{s+\frac{1}{s}}=-\frac{1}{s+\frac{1}{s}}+\frac{1}{(s^{2}+1)(s+\frac{1}{s})}
[/tex]
First of all, is it correct so far?
And how do I continue the simplification from here to be able to use inverse lapace transform?