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ranyai12
Group Title
Help Please!!
Use Laplace transforms to solve the initial value problem
x''+4x'+8x=2e^(t) ; x(0)=0 and x'(0)=4
 one year ago
 one year ago
ranyai12 Group Title
Help Please!! Use Laplace transforms to solve the initial value problem x''+4x'+8x=2e^(t) ; x(0)=0 and x'(0)=4
 one year ago
 one year ago

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ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
@TuringTest
 one year ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
this is what I got so far
 one year ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
after this I am stuck :(
 one year ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
and this is the table I used: http://tutorial.math.lamar.edu/pdf/Laplace_Table.pdf
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
use partial fractions on the \[\frac2{s(s+1)(s+4)}\]part and I think it should be pretty straightforward to do inverse Laplace
 one year ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
but what I do about the 4/s^2+4s
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
factor out the s and that is what I wrote
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
oh I see; same thing though factor out s and do partial fractions
 one year ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
but then you get 4/(s(s+4)>4/s[1/s+4)> 4e^(4t)
 one year ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
is that right?
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
\[\frac4{s^2+4s}=\frac As+\frac B{s+4}\implies A(s+4)+Bs=4\]\[A=1~~~,B=1\]
 one year ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
oh so you do partial fractions with both of them
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
that's what I'd do, yeah
 one year ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
and for the other one it's A(s^2+4s) B(s+1) right
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
\[\mathcal L^{1}\{\frac1s\}=1\]\[\mathcal L^{1}\{\frac1{s+4}\}=e^{4t}\]
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
for the other one it would be...
 one year ago

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

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
\[A(s+1)(s+4)+Bs(s+4)+Cs(s+1)=2\]
 one year ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
then you get As^2+5As+4A+Bs^2+4Bs+Cs^2+C=2
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
not necessary I don't think plug in s=1,4,and 0 and I think you can get the values quickly
 one year ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
so you dont need ti distribute it?
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
not necessarily how did I get the other answers? I'll show you...
 one year ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
i didnt get the answers yet i thiught you had to distribute it then ge tthe answers but now i see that its unnecessary
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
\[\frac4{s^2+4s}=\frac As+\frac B{s+4}\implies A(s+4)+Bs=4\]now plug in\[s=0:A(0+4)+B(0)=4 \implies 4A=4\implies A=1\]
 one year ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
B=2/3 C=1/6
 one year ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
and A=1/2
 one year ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
thats what I got
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
this is the second term:\[\frac4{s^2+4s}=\frac As+\frac B{s+4}\implies A(s+4)+Bs=4\]now plug in\[s=4:A(4+4)+B(4)=4\implies 4B=4\implies B=1\]
 one year ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
I didnt get those values though :(
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
do you see how I got mine?
 one year ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
i mean for the values of the other one
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
that I haven't checked yet, one sec...
 one year ago

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\[A(s+1)(s+4)+Bs(s+4)+Cs(s+1)=2\]\[s=1:3B=2\implies B=\frac23\]\[s=4:12C=2\implies C=\frac16\]\[s=0:4A=2\implies A=\frac12\]yes I guess you are right :)
 one year ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
yay!! so would the final answer be (1/2)(2/3)(e^t)+(1/6)e^(4t)e^(4t)
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
looks good to me :)
 one year ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
it's wrong :(
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
I think you forgot the one from the other term do you have infinite tries?
 one year ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
no I ownly have 4 more tries
 one year ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
but what did I forget
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
ok let me try it again from the top; I went of what you had I should doublecheck everything
 one year ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
ok thanks
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
ok I think I found it...
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
\[x''(0)+4x'+8x=2e^{t}\]\[s^2X(s)sx(0)x'(0)+4sX(s)4x(0)+8X(s)=\frac2{s+1}\]\[(s^2+4s+8)X(s)=\frac2{s+1}+4\]
 one year ago

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you dropped the 8...
 one year ago

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

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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
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
hyperbolic trig functions*
 one year ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
can you please show me because I honestly have no clue
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
\[X(s)={2\over{((s+2)^2+4)(s+1)}}+\frac4{((s+2)^2+4)}\]\[={2\over{((s+2)^2+4)(s+1)}}+2\frac2{((s+2)^2+2^2)}\]the second term is form 19 on the chart in your link
 one year ago

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

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
the other part... you're gonna have to use partial fractions fractions and it looks like it's gonna be a pain
 one year ago

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\[{2\over{((s+2)^2+4)(s+1)}}={As+B\over(s+2)^2+4}+{C\over s+1}\]\[(As+B)(s+1)+C[(s+2)^2+4]=2\]\[s=1:5C=2\implies C=\frac25\]\[s=0:B+\frac{16}5=2\implies B=\frac65\]
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
I think you'll have to get A on your own ;) I'm a bit busy
 one year ago

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I think\[(A+C)s^2=0\implies A=C=\frac25\]but you shouod doublecheck me
 one year ago

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I think that is correct
 one year ago
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