Question

inverse laplace transform of

(s^2-3s+1)/(s^3+s^2-2s)

Answer 1

the denominator will be s(s+2)(s-1), but im not sure what to do with the numerator?

Answer 2

tried partial fraction ?

Answer 3

does it split into 3 fractions though? im really trying to remember partial fractions its been so long

Answer 4

yes, it does,...
\(\dfrac{s^2-3s+1}{s(s+2)(s-1)}=\dfrac{A}{s}+\dfrac{B}{s+2}+\dfrac{C}{s-1}=\)

Answer 5

know how to find A,B,C ?

Answer 6

umm well to find A you would divide s^2+s-2 into s^2-3s+1?? that doesnt seem right though

Answer 7

first, do this,
s^2-3s+1 = A(....) + B (.....) +C (.....)
can you fill the blanks ? simple algebra

Answer 8

actually i dont remember this at all, are you talking about just putting the s^2 into the first, and -3s into the second and 1 into the third??

Answer 9

no...i was asking you to make a common denominator for left side,
like
w/x+y/z = (wz+xy)/xz

Answer 10

FOR RIGHT SIDE***

Answer 11

\(\dfrac{s^2-3s+1}{s(s+2)(s-1)}=\dfrac{A}{s}+\dfrac{B}{s+2}+\dfrac{C}{s-1} \\ \implies s^2-3s+1= A (s+2)(s-1)+Bs(s-1)+Cs(s+2) \)
did you get this ???

Answer 12

ohhh yeah, thats what i was saying with s^2+s-2... for A, I guess since you didn't say anything I was confused... but yes, im totally with you.

Answer 13

to find A, put s=0
what u get ?

Answer 14

i just noticed that i typed it in wrong... it is s^2-3s-1.

but to find A, are u saying plug in 0 for s and you get -2A and you solve setting that equal to s^2-3s-1... so then just divide by -2

Answer 15

wait then it would be 1/2

Answer 16

ok, then
\(s^2-3s-1= A (s+2)(s-1)+Bs(s-1)+Cs(s+2)\)
A= 0
-1 =A (-2)
yes, A =1/2 correct,
now you know how can you find B ?

Answer 17

well first, why did you pick s=0?? you can't use that for B now though

Answer 18

look at this equation closely
\(s^2-3s-1= A (s+2)(s-1)+Bs(s-1)+Cs(s+2)\)

this is true for every value of 's'
now if i want to find A , what value of 's' can i put so that i will not get B and C in calculations ?

Answer 19

gotcha, so to find b we use -2 and to find c we use 1.

Answer 20

absolutely correct!

Answer 21

k, i will get those.

Answer 22

B=3/2 and C=1

Answer 23

c=-1

Answer 24

correct :)
now take the inverse LT of individual terms,
\(\dfrac{s^2-3s+1}{s(s+2)(s-1)}=\dfrac{1}{2s}+\dfrac{3}{2(s+2)}-\dfrac{1}{s-1}\)
can you ?

Answer 25

well then f(x) = 1/2 + 3/2e^-2x - e^-x

*just followed the table of inverse LT.

Answer 26

yes, i get the same.....
i would have used variable 't' instead..

Answer 27

is there any particular reason why use t instead?

Answer 28

habit....
if F (x) = (....) then use 'x'
if F (t) = (....) then use 't'
else use any variable...