ranyai12
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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ranyai12
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@TuringTest
ranyai12
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this is what I got so far
ranyai12
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after this I am stuck :(
TuringTest
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use partial fractions on the \[\frac2{s(s+1)(s+4)}\]part and I think it should be pretty straightforward to do inverse Laplace
ranyai12
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but what I do about the 4/s^2+4s
TuringTest
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factor out the s and that is what I wrote
TuringTest
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oh I see; same thing though
factor out s and do partial fractions
ranyai12
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but then you get 4/(s(s+4)-->4/s[1/s+4)--> 4e^(-4t)
ranyai12
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is that right?
TuringTest
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\[\frac4{s^2+4s}=\frac As+\frac B{s+4}\implies A(s+4)+Bs=4\]\[A=1~~~,B=-1\]
ranyai12
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oh so you do partial fractions with both of them
TuringTest
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that's what I'd do, yeah
ranyai12
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and for the other one it's A(s^2+4s) B(s+1) right
TuringTest
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\[\mathcal L^{-1}\{\frac1s\}=1\]\[-\mathcal L^{-1}\{\frac1{s+4}\}=-e^{-4t}\]
TuringTest
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for the other one it would be...
TuringTest
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\[\frac2{s(s+1)(s+4)}=\frac As+\frac B{s+1}+\frac C{s+4}\]
TuringTest
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\[A(s+1)(s+4)+Bs(s+4)+Cs(s+1)=2\]
ranyai12
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then you get As^2+5As+4A+Bs^2+4Bs+Cs^2+C=2
ranyai12
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right
TuringTest
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not necessary I don't think
plug in s=-1,-4,and 0 and I think you can get the values quickly
ranyai12
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so you dont need ti distribute it?
TuringTest
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not necessarily
how did I get the other answers?
I'll show you...
ranyai12
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i didnt get the answers yet i thiught you had to distribute it then ge tthe answers but now i see that its unnecessary
TuringTest
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\[\frac4{s^2+4s}=\frac As+\frac B{s+4}\implies A(s+4)+Bs=4\]now plug in\[s=0:A(0+4)+B(0)=4 \implies 4A=4\implies A=1\]
ranyai12
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B=-2/3
C=1/6
ranyai12
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and A=1/2
ranyai12
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thats what I got
TuringTest
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this is the second term:\[\frac4{s^2+4s}=\frac As+\frac B{s+4}\implies A(s+4)+Bs=4\]now plug in\[s=-4:A(-4+4)+B(-4)=4\implies -4B=4\implies B=-1\]
ranyai12
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I didnt get those values though :(
TuringTest
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do you see how I got mine?
ranyai12
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i mean for the values of the other one
TuringTest
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that I haven't checked yet, one sec...
TuringTest
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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 :)
ranyai12
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yay!! so would the final answer be (1/2)-(2/3)(e^-t)+(1/6)e^(-4t)-e^(-4t)
TuringTest
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looks good to me :)
ranyai12
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it's wrong :(
TuringTest
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I think you forgot the one from the other term
do you have infinite tries?
ranyai12
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no I ownly have 4 more tries
ranyai12
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but what did I forget
ranyai12
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only*
TuringTest
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ok let me try it again from the top; I went of what you had
I should double-check everything
ranyai12
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ok thanks
TuringTest
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ok I think I found it...
TuringTest
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\[x''(0)+4x'+8x=2e^{-t}\]\[s^2X(s)-sx(0)-x'(0)+4sX(s)-4x(0)+8X(s)=\frac2{s+1}\]\[(s^2+4s+8)X(s)=\frac2{s+1}+4\]
TuringTest
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you dropped the 8...
TuringTest
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unfortunately this is going to make partial fractions a real pain...
TuringTest
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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
TuringTest
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hyperbolic trig functions*
ranyai12
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can you please show me because I honestly have no clue
TuringTest
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\[X(s)={2\over{((s+2)^2+4)(s+1)}}+\frac4{((s+2)^2+4)}\]\[={2\over{((s+2)^2+4)(s+1)}}+2\frac2{((s+2)^2+2^2)}\]the second term is form 19 on the chart in your link
TuringTest
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\[\mathcal L^{-1}\{\frac2{((s+2)^2+2^2)}\}=e^{-2t}\sin(2t)\]
TuringTest
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the other part... you're gonna have to use partial fractions fractions and it looks like it's gonna be a pain
TuringTest
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\[{2\over{((s+2)^2+4)(s+1)}}={As+B\over(s+2)^2+4}+{C\over s+1}\]\[(As+B)(s+1)+C[(s+2)^2+4]=2\]\[s=-1:5C=2\implies C=\frac25\]\[s=0:B+\frac{16}5=2\implies B=-\frac65\]
TuringTest
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I think you'll have to get A on your own ;)
I'm a bit busy
TuringTest
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I think\[(A+C)s^2=0\implies A=-C=-\frac25\]but you shouod double-check me
TuringTest
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I think that is correct