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

- anonymous

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- anonymous

@TuringTest

- anonymous

this is what I got so far

##### 1 Attachment

- anonymous

after this I am stuck :(

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## More answers

- anonymous

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

- TuringTest

use partial fractions on the \[\frac2{s(s+1)(s+4)}\]part and I think it should be pretty straightforward to do inverse Laplace

- anonymous

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

- TuringTest

factor out the s and that is what I wrote

- TuringTest

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

- anonymous

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

- anonymous

is that right?

- TuringTest

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

- anonymous

oh so you do partial fractions with both of them

- TuringTest

that's what I'd do, yeah

- anonymous

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

- TuringTest

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

- TuringTest

for the other one it would be...

- TuringTest

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

- TuringTest

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

- anonymous

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

- anonymous

right

- TuringTest

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

- anonymous

so you dont need ti distribute it?

- TuringTest

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

- anonymous

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

\[\frac4{s^2+4s}=\frac As+\frac B{s+4}\implies A(s+4)+Bs=4\]now plug in\[s=0:A(0+4)+B(0)=4 \implies 4A=4\implies A=1\]

- anonymous

B=-2/3
C=1/6

- anonymous

and A=1/2

- anonymous

thats what I got

- TuringTest

this is the second term:\[\frac4{s^2+4s}=\frac As+\frac B{s+4}\implies A(s+4)+Bs=4\]now plug in\[s=-4:A(-4+4)+B(-4)=4\implies -4B=4\implies B=-1\]

- anonymous

I didnt get those values though :(

- TuringTest

do you see how I got mine?

- anonymous

i mean for the values of the other one

- TuringTest

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

- TuringTest

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

- anonymous

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

- TuringTest

looks good to me :)

- anonymous

it's wrong :(

- TuringTest

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

- anonymous

no I ownly have 4 more tries

- anonymous

but what did I forget

- anonymous

only*

- TuringTest

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

- anonymous

ok thanks

- TuringTest

ok I think I found it...

- TuringTest

\[x''(0)+4x'+8x=2e^{-t}\]\[s^2X(s)-sx(0)-x'(0)+4sX(s)-4x(0)+8X(s)=\frac2{s+1}\]\[(s^2+4s+8)X(s)=\frac2{s+1}+4\]

- TuringTest

you dropped the 8...

- TuringTest

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

- TuringTest

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

hyperbolic trig functions*

- anonymous

can you please show me because I honestly have no clue

- TuringTest

\[X(s)={2\over{((s+2)^2+4)(s+1)}}+\frac4{((s+2)^2+4)}\]\[={2\over{((s+2)^2+4)(s+1)}}+2\frac2{((s+2)^2+2^2)}\]the second term is form 19 on the chart in your link

- TuringTest

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

- TuringTest

the other part... you're gonna have to use partial fractions fractions and it looks like it's gonna be a pain

- TuringTest

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

- TuringTest

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

- TuringTest

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

- TuringTest

I think that is correct

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