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this is what I got so far

after this I am stuck :(

and this is the table I used: http://tutorial.math.lamar.edu/pdf/Laplace_Table.pdf

but what I do about the 4/s^2+4s

factor out the s and that is what I wrote

oh I see; same thing though
factor out s and do partial fractions

but then you get 4/(s(s+4)-->4/s[1/s+4)--> 4e^(-4t)

is that right?

\[\frac4{s^2+4s}=\frac As+\frac B{s+4}\implies A(s+4)+Bs=4\]\[A=1~~~,B=-1\]

oh so you do partial fractions with both of them

that's what I'd do, yeah

and for the other one it's A(s^2+4s) B(s+1) right

\[\mathcal L^{-1}\{\frac1s\}=1\]\[-\mathcal L^{-1}\{\frac1{s+4}\}=-e^{-4t}\]

for the other one it would be...

\[\frac2{s(s+1)(s+4)}=\frac As+\frac B{s+1}+\frac C{s+4}\]

\[A(s+1)(s+4)+Bs(s+4)+Cs(s+1)=2\]

then you get As^2+5As+4A+Bs^2+4Bs+Cs^2+C=2

right

not necessary I don't think
plug in s=-1,-4,and 0 and I think you can get the values quickly

so you dont need ti distribute it?

not necessarily
how did I get the other answers?
I'll show you...

B=-2/3
C=1/6

and A=1/2

thats what I got

I didnt get those values though :(

do you see how I got mine?

i mean for the values of the other one

that I haven't checked yet, one sec...

yay!! so would the final answer be (1/2)-(2/3)(e^-t)+(1/6)e^(-4t)-e^(-4t)

looks good to me :)

it's wrong :(

I think you forgot the one from the other term
do you have infinite tries?

no I ownly have 4 more tries

but what did I forget

only*

ok let me try it again from the top; I went of what you had
I should double-check everything

ok thanks

ok I think I found it...

you dropped the 8...

unfortunately this is going to make partial fractions a real pain...

hyperbolic trig functions*

can you please show me because I honestly have no clue

\[\mathcal L^{-1}\{\frac2{((s+2)^2+2^2)}\}=e^{-2t}\sin(2t)\]

I think you'll have to get A on your own ;)
I'm a bit busy

I think\[(A+C)s^2=0\implies A=-C=-\frac25\]but you shouod double-check me

I think that is correct