Question

Help Please!!

Use Laplace transforms to solve the initial value problem

x''+4x'+8x=2e^(-t) ; x(0)=0 and x'(0)=4

Answer 1

@TuringTest

Answer 2

this is what I got so far

Answer 3

after this I am stuck :(

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

factor out the s and that is what I wrote

Answer 8

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

Answer 9

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

Answer 10

is that right?

Answer 11

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

Answer 12

oh so you do partial fractions with both of them

Answer 13

that's what I'd do, yeah

Answer 14

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

Answer 15

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

Answer 16

for the other one it would be...

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

right

Answer 21

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

Answer 22

so you dont need ti distribute it?

Answer 23

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

Answer 24

i didnt get the answers yet i thiught you had to distribute it then ge tthe answers but now i see that its unnecessary

Answer 25

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

Answer 26

B=-2/3
C=1/6

Answer 27

and A=1/2

Answer 28

thats what I got

Answer 29

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

Answer 30

I didnt get those values though :(

Answer 31

do you see how I got mine?

Answer 32

i mean for the values of the other one

Answer 33

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

Answer 34

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

Answer 35

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

Answer 36

looks good to me :)

Answer 37

it's wrong :(

Answer 38

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

Answer 39

no I ownly have 4 more tries

Answer 40

but what did I forget

Answer 41

only*

Answer 42

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

Answer 43

ok thanks

Answer 44

ok I think I found it...

Answer 45

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

Answer 46

you dropped the 8...

Answer 47

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

Answer 48

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

Answer 49

hyperbolic trig functions*

Answer 50

can you please show me because I honestly have no clue

Answer 51

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

Answer 52

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

Answer 53

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

Answer 54

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

Answer 55

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

Answer 56

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

Answer 57

I think that is correct