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Help Please!! Use Laplace transforms to solve the initial value problem x''+4x'+8x=2e^(-t) ; x(0)=0 and x'(0)=4

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this is what I got so far
1 Attachment
after this I am stuck :(

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Other answers:

and this is the table I used:
use partial fractions on the \[\frac2{s(s+1)(s+4)}\]part and I think it should be pretty straightforward to do inverse Laplace
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...
\[\frac2{s(s+1)(s+4)}=\frac As+\frac B{s+1}+\frac C{s+4}\]
then you get As^2+5As+4A+Bs^2+4Bs+Cs^2+C=2
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...
i didnt get the answers yet i thiught you had to distribute it then ge tthe answers but now i see that its unnecessary
\[\frac4{s^2+4s}=\frac As+\frac B{s+4}\implies A(s+4)+Bs=4\]now plug in\[s=0:A(0+4)+B(0)=4 \implies 4A=4\implies A=1\]
B=-2/3 C=1/6
and A=1/2
thats what I got
this is the second term:\[\frac4{s^2+4s}=\frac As+\frac B{s+4}\implies A(s+4)+Bs=4\]now plug in\[s=-4:A(-4+4)+B(-4)=4\implies -4B=4\implies B=-1\]
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...
\[A(s+1)(s+4)+Bs(s+4)+Cs(s+1)=2\]\[s=-1:-3B=2\implies B=-\frac23\]\[s=-4:12C=2\implies C=\frac16\]\[s=0:4A=2\implies A=\frac12\]yes I guess you are right :)
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
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...
write is as a complete square\[((s+2)^2+4)X(s)=\frac2{s+1}+4\]you should wind up with some hyperbolic I think
hyperbolic trig functions*
can you please show me because I honestly have no clue
\[X(s)={2\over{((s+2)^2+4)(s+1)}}+\frac4{((s+2)^2+4)}\]\[={2\over{((s+2)^2+4)(s+1)}}+2\frac2{((s+2)^2+2^2)}\]the second term is form 19 on the chart in your link
\[\mathcal L^{-1}\{\frac2{((s+2)^2+2^2)}\}=e^{-2t}\sin(2t)\]
the other part... you're gonna have to use partial fractions fractions and it looks like it's gonna be a pain
\[{2\over{((s+2)^2+4)(s+1)}}={As+B\over(s+2)^2+4}+{C\over s+1}\]\[(As+B)(s+1)+C[(s+2)^2+4]=2\]\[s=-1:5C=2\implies C=\frac25\]\[s=0:B+\frac{16}5=2\implies B=-\frac65\]
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

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