inverse laplace transform, example#4, with partial fraction

[@not_potaytoes_hobbit]

I had never seen this method for solving partial fractions in the 4 years I have spent in college and it's a life saver, thank you!

[@ahmedmalakai9753]

You probably dont give a shit but does someone know a method to log back into an Instagram account?? I was stupid forgot the account password. I love any tips you can offer me

[@christianelzey9703]

BEST explanation I have seen so far. Thank you very much.

[@joel6253]

this is going to save my butt on my signals and systems test tommorow. thank you

[@DrQuatsch]

Yeah, I did it in a completely different way! I completed the square at the bottom: s^2 + 2s - 3 = (s + 1)^2 - 2^2. Then I split the inverse laplace transform into: L^-1{(s + 1)/((s+1)^2 - 2^2)} + L^-1{3/((s+1)^2 - 2^2)}. The s+1 gives us a common factor of e^-t, so it becomes e^-5 * (L^-1{s/(s^2 - 2^2)} + 3/2 L^-1{2/(s^2 - 2^2}) = e^-t * (cosh(2t) + 3/2 sinh(2t)). If you substitute the cosh and sinh for their e-power forms, it results in the same answer, obviously.

[@NKShahariar]

I did it in the same way as you did. Thanks for the cross check

[@maxw5667]

Thank you I couldn't find any comprehensive explanation of the inverse laplace apart from these videos

[@Rstshclfrd]

Thanks dude Youre really good at explaining in a good pace Really helped me comprehend

[@blackpenredpen]

otheamazing my pleasure!

[@trymoremuchemwa9616]

Hie sir ,you ae too much ,i like your lectures

[@brayschway6746]

This was more than helpful. Thanks a ton!

[@qureshisiddig9274]

You're best of the best of all

[@jkuebler89]

Oh my gosh this is why you are my favorite. Took something scary and made digestible.

[@husseinqabbout6855]

Omg i love you!

[@JacobPorterLadder]

where were these videos 5 years ago when i was in school. studying for my FE and these are a Lifesaver (make sure you take your FE before you graduate kids)

[@lawtraf8008]

What FE ?

[@JacobPorterLadder]

@@lawtraf8008 electrical

[@sanaarshad470]

amazing

[@luc6457]

thx saved me an hour to study this topic

[@johnkyingilisi4395]

rather a "masochistic "way to find couple of constants. try partial fraction, less painful.

[@BillFye]

Is it wrong to use completing the square with this problem? Do you have to use partial fraction?

[@soomaseirt3615]

Hi

[@pinkynicolettecabalog2339]

thank you for this! based from inverse laplace table

[@emilyroberts108]

I used completing the square on the denominator and got a different answer: e^(-t)*[cos(2t)+3sin(2t)] - did I do something wrong?

[@ov3rkill]

Yes, you will arrive at the same answer but it'll be a long and tedious solution since you'll be using trigonometric identities a lot until you will end up at e.

[@mccoy5452]

What concept did you apply on getting the numerator of the right side?

[@hizmoryyii1833]

what if the denominator for both is equal?

[@sanashaikh5429]

Excellent job

[@Danish13012000]

What if you have 3 terms as the denominator? Do you do faction over fraction over fraction?

[@johndaveleones69420]

Thank you mate. Now I can make my plates. 😂❤

[@golmolenahisidhibaatein6180]

Hi, perfect. Is laplace inverse of (2/p^2-2p)= 2e^2t. or we apply Convolution theorem of inverse laplace. or 2 L'(1/(p-1)^2-1)=2/1 (Sinht). sir pls reply.

[@kondojuma5489]

Good shortly and clear

[@helloitsme7553]

Can you do this with completing the square on the bottom and using the translation theorem please

[@helloitsme7553]

Never mind that'd be with complex numbers

[@erickrobles6365]

why with complex numbers only?

[@martinepstein9826]

You don't need complex numbers, you need sinh and cosh.

[@sidraduhoki5893]

You’re perfect! Thanks :)

[@soomaseirt3615]

Thanks for helping, and deliver the issue of the issue if it executed F (s) = s-1 / s3 + 2s2 + s

[@abhishekchakrabarty2930]

I did it with another concept and my answer comes epow(-t)[cos2t+1.5sin2t] Is it right?

[@vo8263]

This guy just covered a partial fraction that will take 4lines with zero lines

[@orionclachar5094]

that shit blew my mind no cap

[@notstemmajor357]

I fw the vision lets link + build

[@pixisilva]

És uma lenda meu tropa , smp aqui

[@zahraaabbas9902]

Thank you soooo much

[@merigonmeri]

Wait...how If we substitute s = 1 to get B as 0 to grt A value...then we get A = 1 How is it 5/4 what

[@narsing1826]

What a shortttttt cuttttt

[@briendamathhatter816]

I got an equivalent answer: (e^-t)(cosh(2t)+(3/2)sinh(2t)) I calculatored it with : (e^-t)(cosh(2t)+(3/2)sinh(2t)) = (5/4)e^t-(1/4)e^(-3t)

[@akashlenka4068]

Yeah bro got the same answer

[@gamaslazuardi330]

Woww soo fast, thanks

[@فيصلعيسى-ف3ش]

أسطورة انت ي رجل

[@ghazalbehzad6135]

khodaish damet garm :))))

[@sohuilee1936]

thanks man!

[@ameermizory4170]

plz ,i need inverse Laplace for (3/(s-2)^2 +25)

[@carultch]

An irreducible quadratic denominator with a constant numerator, is the Laplace transform of an exponential multiplied by a sine wave. From the standard table of Laplace transforms: £{e^(a*t) * sin(b*t)} = b/((s - a)^2 + b^2) To get your expression to look like this, we identify that a=2 and b=5. We then need to multiply by 1 in a fancy way (i.e. 5/5), to make your numerator also equal 5. 3/((s-2)^2 +25) * 5/5 = 3/5 * 5/(s-2)^2 +25) Pull the 3/5 out in front as a constant, and evaluate, and we have our solution: 3/5 * £^(-1) {5/(s-2)^2 +25)} = 3/5 * e^(2*t) * sin(5*t) £^(-1) {3/(s-2)^2 +25)} = 3/5 * e^(2*t) * sin(5*t)

[@АйманШегетаева-ы1ш]

frac{2}{3}

[@blackmamba8600]

thank you bro

[@NSMQ1998]

I like your accent😍

[@klevzor]

thank you so much

[@duhanassif1364]

Thank you

[@rizzamaymerencilla6903]

💕

[@saurabhjha752]

i didn't understood that how s square plus 2 equals to s-1?

[@darkcnotion]

Saurabh Jha he's using a trick

[@TriggeredBicep]

Im ready for my Final....i hope

[@karanveersingh1234]

blue pen?!?!?!

[@marym.1149]

thannkkkk youuu sooo much haha

[@hobyedukasi248]

translite indonesia ??

[@senthilmurugan6195]

24000 GBP uni tuition and failed after attending 2 hours class. 3 min youtube video and i have completely understood the process.

[@oshanehamilton3592]

same here

[@chaitanyachollangi1330]

Tq bro

[@pumpmanagementsoftware]

Thanks

[@blackpenredpen]

My pleasure!

[@jorgehabib1815]

i love you

[@chandanmehta9156]

tq

[@blackpenredpen]

what's tq?

[@martinepstein9826]

+blackpenredpen First time I've seen it but I read it like ty = thank you tq = than q (say it fast)

[@rajeshrs1226]

This will happen When your phone charge is at 1 % and you want to make videos for your subscribers😂😂

[@kumitatataaa2945]

thank you T_T

[@mimisel5367]

Thanks