anonymous
  • anonymous
find inverse laplace transform of F(s)= (s^2+2s+5/(s^2(s+1))
Mathematics
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
katieb
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
note
anonymous
  • anonymous
'note' sorry this is my first time using this what do you mean by that?
anonymous
  • anonymous
Im just bookmarking this question so that I can return to it later. :D Id want to know the answer too

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
oh ok sounds good thanks
lgbasallote
  • lgbasallote
your question is \[\huge \mathcal L ^{-1} \left \{ s^2 + 2s + \frac 5{s^2(s+1)}\right \}\]
lgbasallote
  • lgbasallote
??
lgbasallote
  • lgbasallote
@ragman you there?
anonymous
  • anonymous
hey sorry about the late reply but my question is actually (s^2+22+5) / (s^2(s+1) sorry i forgot the other )
anonymous
  • anonymous
@lgbasallote
lgbasallote
  • lgbasallote
uhh s^2 + 22 + 5??
anonymous
  • anonymous
dang i keep doing typos 2s not 22 so yeah its (s^2+2s+5)
lgbasallote
  • lgbasallote
so.. \[\huge \mathcal L ^{-1} \left \{ \frac{s^2 + 2s + 5}{s^2(s+1)} \right \}\]
anonymous
  • anonymous
yup perfect
hartnn
  • hartnn
can u do partial fractions ??
anonymous
  • anonymous
@hartnn yeah thats what i was doing but for some reason i could figure out the other two variable maybe im solving it incorrectly but i got -3 for A and 5 for B
lgbasallote
  • lgbasallote
where's C?
hartnn
  • hartnn
those are correct, so u need only C right ? where u took C above s+1
lgbasallote
  • lgbasallote
your P.F. should be \[\huge \frac As + \frac B{s^2} + \frac C{s+1}\]
lgbasallote
  • lgbasallote
so you have \[\implies As(s+1) + B(s+1) + Cs^2 = s^2 + 2s + 5\]
anonymous
  • anonymous
oh ok so thats where i messed up i thought that where C is it should be (Cs+D) so i was trying to solve for both of those variables
hartnn
  • hartnn
nopes only C, if it were s^2+1 u would have taken Cx+D
hartnn
  • hartnn
so i think u can get C now, right?
anonymous
  • anonymous
oh i see now and yeah i believe i can get C now trying it right now
lgbasallote
  • lgbasallote
i think it will change your A value too
lgbasallote
  • lgbasallote
so you best check
hartnn
  • hartnn
great tell us what u get C as....we all will verify that. A is correct.
lgbasallote
  • lgbasallote
nope nevermind..it stays the same
anonymous
  • anonymous
ok sounds good got B=5 finding A now
anonymous
  • anonymous
A=-3 and C=4 is that what you guys got?
hartnn
  • hartnn
yup, same.Good work.
lgbasallote
  • lgbasallote
A =-3?
lgbasallote
  • lgbasallote
yeah that's right
anonymous
  • anonymous
ok cool i was about to write out how i got it lol
lgbasallote
  • lgbasallote
everything's right
anonymous
  • anonymous
and now i just plug back all the variables right
hartnn
  • hartnn
yup, and use inverse laplace formulas.
lgbasallote
  • lgbasallote
yep find \[\huge \mathcal L ^{-1} \left \{ -\frac 3s + \frac 5s + \frac 4{s+1} \right \}\]
lgbasallote
  • lgbasallote
uhh that should be 5/s^2
anonymous
  • anonymous
ok sorry im just learning this but how do you go about doing that im trying to find the L^-1 formula but my notes arent really helping
lgbasallote
  • lgbasallote
here's a hint: \[\huge \mathcal L \{ a \} = \frac as\] \[\huge \mathcal L \{ t^n\} = \frac{n!}{s^{n+1}}\] \[\huge \mathcal L \{e^{at} \} = \frac{a}{s -a}\] does that help?
lgbasallote
  • lgbasallote
uhhh wait... wrong formula for last one
lgbasallote
  • lgbasallote
\[\huge \mathcal L \{e^{at} \} = \frac{1}{s-a}\] better
anonymous
  • anonymous
yes thats very helpful but im getting confused with what n should be in the middle equations
lgbasallote
  • lgbasallote
n is the exponent
anonymous
  • anonymous
so it would be 2?
lgbasallote
  • lgbasallote
for example: \[\huge \mathcal L \{ t^2 \} = \frac{2!}{s^{2+1}} \implies \frac{2}{s^3}\]
anonymous
  • anonymous
thats whats a lil confusing to me because on the denominator it would have to be n=1 so that it would equal s^2 but on the numerator 1! equals 1 right ?
hartnn
  • hartnn
yes thats true, so u will have 5t^1= 5t ....isn't it ? u can verify that u should get final answer as 5t-3+4e^(-t) did u get that?
anonymous
  • anonymous
yup thats what i got thanks
hartnn
  • hartnn
welcome :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.