Inverse Laplace transform question:
$$Y(t)=\frac{\frac{1}{s+1}+1}{s^2-2s+2}$$
It came from the IVP $y''-2y'+2y=e^{-t}$ I just cannot seem to reduce it into a doable inverse

Question

Inverse Laplace transform question:
$$Y(t)=\frac{\frac{1}{s+1}+1}{s^2-2s+2}$$
It came from the IVP $y''-2y'+2y=e^{-t}$  I just cannot seem to reduce it into a doable inverse

perl · Answer

you have two equal signs in your diff. equation

fibonaccichick666 · Answer

@ganeshie8, @Zarkon, @Mashy Do y'all have any ideas? I've tried partial fractions but it didn't work

fibonaccichick666 · Answer

oops, first should be a minus

perl · Answer

you can edit your question

fibonaccichick666 · Answer

better?

fibonaccichick666 · Answer

@perl , any ideas on the simplification though?

anonymous · Answer

$$\begin{align*}Y(s)&=\frac{\dfrac{1}{s+1}+1}{s^2-2s+2}\\
&=\frac{\dfrac{1+s+1}{s+1}}{(s-1)^2+1}\\
&=\frac{s+2}{(s+1)((s-1)^2+1)}\end{align*}$$
Partial fractions from here?

fibonaccichick666 · Answer

yea, that's what I tried, but I got a=0 contradicting with a= -2

fibonaccichick666 · Answer

did I just mess up the algebra?

anonymous · Answer

$$\begin{align*}
\frac{s+2}{\cdots}&=\frac{A}{s+1}+\frac{Bs+C}{(s-1)^2+1}\\
s+2&=A(s^2-2s+2)+(Bs+C)(s+1)\\
&=(A+B)s^2+(-2A+B+C)s+2A+C
\end{align*}$$
which gives the system
$$\begin{cases}
A+B=0\
-2A+B+C=1\
2A+C=2
\end{cases}~~\implies~~A=\frac{1}{5},~B=-\frac{1}{5},~C=\frac{8}{5}$$

fibonaccichick666 · Answer

oh... yep

fibonaccichick666 · Answer

I forgot you have to do a Bs+C...ughhh

fibonaccichick666 · Answer

nooooo wonder

anonymous · Answer

Mmm yeah it's a quadratic factor in disguise :P

fibonaccichick666 · Answer

I'm good from here.  thanks

anonymous · Answer

You're welcome