14. Inverse Laplace Transforms | Problem#2 | Most Important

[@ritikashandilya8886]

thx a lot sir...i was very confused.you just explained it very well and also gave me idea to atempt in exam

[@kiranteja3181]

Thank'u Mr. Manoj, you have been a great help.

[@shipraarora5559]

Thanks sir aapnay bohat easy method karvaya mujay is Sam mai pahlay doubt that aab clear Ho Gaya

[@muskankhan13921]

Thx I was just searching this question...

[@anushkasingh9686]

thank u sar aapki videos mere bahut helpful rahi hai mujhe Laplace theorem bilkul samajh mein Nahin a raha tha aap ki videos dekh kar mujhe bahut kuch help Mili hai thank you so much sir thank you thank u 🤗

[@dinkark98]

Thanks a lot, sir for teaching me this Approach............

[@elaf7092]

Very helpful,sir

[@PIYUSHSHARMA-kw5hi]

thnx a lot sir

[@swatitripathi9864]

sir your videos are really helpful, through your videos I am able to learn this chapter very quickly and it's helping me even while revision. The kind of questions you cover is seriously very important and mostly I encounter exactly those types of questions in my test. Before doing maths textbook I first see your videos then solve the book as it becomes very easy to solve after that. Thank you!

[@ManojKumarSahoo]

Good Luck

[@meghajoshi1424]

Sir in important questions se bahut help.mil raha h

[@RahulKumar-zo9qo]

Nice sir acha se samjh me a gay sir thanks sir aapki video bahut achaa laga sir

[@devashishbiswas1598]

outstanding class.. sir

[@meghajoshi1424]

Thankyou so much very very useful video

[@santoshvishawkarm7640]

Thanks for teaching

[@ashimdas2207]

It was awesome... But it was very tough also if u have any other method for this question then plzz make an another for this question..

[@Vishal-J97]

Sir... Aaj mujhe.. Ye question ka ansr jaake mila... Maine Saare jagha... Almost search kiya.. But koi channel me ni.. Mila... But sir aapke subscribers aur hone chhahiye... Thnx.. Sir... Mai iss question ka solution pura 1 din tak dhunda par ni mila... I will share it sir..! Aise questions ka solution kaafi help krte hai sir... Thnx once again... Keep it up..!

[@shanidhatiria]

Same 🙏

[@PrashantSingh-ld7jq]

Thank you so much sir You explanation is very good

[@bharathviswath2186]

thanks for a really good problem and explanation sir

[@BROKENHEART-ze2cq]

Sir ji you are my ramanujan✍️🙏🙏🙏🙏🙏

[@ashadnasim8103]

thanksir for all the videos .You helped me a lot

[@uttamchanduttamchand6755]

Very Gud teaching style sir.ur lec are very helpful. Thanx for these lec 😊

[@ManojKumarSahoo]

Good Luck

[@nikiraj8965]

Thankuu soo much sir

[@marwamohammed7849]

Thank you very much 🌹

[@ahmedtalib7627]

thx u very much

[@subhashmandal5027]

Thanks 🙏 sir

[@imrulkayes894]

Thanks👍

[@028harshkumar7]

Thank u sir

[@huzaifamuhammadhamid1984]

Nice explain ✨👍

[@fazilraheman6147]

Thank you sir

[@Ani_Prajapati]

bhot helpful 😆😆

[@arthtripathi6663]

Sir ji aap hi hamara uddhaar Kar rhe Hai thanks sir for helping us bsc ke baad aapko yaad karenge apna koi no zaroor dijiyega

[@allenjonesstyles6112]

U can easily solve it using convolution theorem

[@mohsinstune51]

Respect boss 💙💙

[@somalrudra5805]

Sir, for sin(at) the numerator also has to have an 'a' if denominator is having 'a'

[@gdkzs6707]

That's why he multiplied denominator with a

[@pK_bihar99]

Make a video on convolution of laplace equations..

[@ManojKumarSahoo]

Please go to playlist of Laplace Transform

[@PrabhatSingh-kk9dx]

Thanks sir

[@prinzekhasim3740]

Sir aapki voice bahut dheemi hai .. Explanation bahut accha hai thoda volume jyada kardo..

[@mukeshsingh925]

by using convolution theorem please solve it s/(s^2-a^2)^3

[@devashishbiswas1598]

sir.. plz share more questions about this topic

[@niharika3237]

Thnk you🥺

[@musicismyfuel4085]

you should make a facebook group for all to discuss questions.. That'd be great! A lot of us even though we try questions, we can not verify its corrrectness.

[@upendrasivalanka9183]

Super explanation sir

[@holyshit922]

If we are limited to real domain, the other way is convolution

[@shaik.wajedahamad3801]

Nice sir...

[@mdsaddam-em9hz]

Thanx sir

[@factmotive4773]

Nice sir

[@mr_nath07]

So nice sir

[@ayushchoudhary8702]

Sir plz Ye question aap convolution theorem se solve kr do

[@kareemabd8087]

Thanks

[@pkr7306]

Best expanded

[@sanatansahoolifestory2083]

THNK U SIR

[@Creativedimag10M]

Concept is amazing but voice is to slow 😢.

[@Mr.M卂NJ]

👍

[@NarendraKumar-lz4fh]

👍Thanque sir👌👌👌👌👌

[@gdkzs6707]

If voice was bit louder then it would be more helpful.

[@vaishnavishirbhate3708]

Sir laplace inverse of 1/s(s^2+4)

[@techmaster6012]

Its very easy u hve to 1st calculate the inverse of 1/s`2+4 then the term comes in the proprty of division....then calc. The value

[@kaushikagrawal6565]

Sir last me 1/s^2+a^2 ka inverse 1/a.sin(at) kaise hoga?

[@ajaysaw2716]

1/s^2 (s^2+a^2)

[@ajaysaw2716]

Answer

[@carultch]

@@ajaysaw2716 Given: Laplace inverse of 1/(s^2 * (s^2 + a^2)) Set up partial fraction expansion: A/s^2 + B/s + (C*s + D)/(s^2 + a^2) Use Heaviside coverup to find A, at s=0: A = 1/(coverup * (0^2 + a^2)) = 1/a^2 Multiply by 1 instance of s, to partially clear the fractions: 1/(s*(s^2 + a^2)) = 1/a^2/s + B + (C*s^2 + D*s)/(s^2 + a^2) Take the limit as s goes to infinity, to set up one of our equations to solve for constants: 0 = B + C Let s = 1, and s=-1, to set up two more equations: s=1: 1/(1^2*(1 + a^2)) = 1/a^2/1^2 + B/1 + (C*1 + D)/(1^2 + a^2) 1/(1 + a^2) = 1/a^2 + B + (C + D)/(1 + a^2) 1 = 1/a^2 + 1 + B*(1 + a^2) + C + D 0 = 1/a^2 + B*(1 + a^2) + C + D s=-1: 1/((-1)^2*((-1)^2 + a^2)) = 1/a^2/(-1)^2 + B/(-1) + (C*(-1) + D)/(-1)^2 + a^2) 1/(1 + a^2) = 1/a^2 - B + (-C + D)/(1 + a^2) 1 = 1 + 1/a^2 - B*(1 + a^2) + (-C + D) 0 = 1/a^2 - B*(1 + a^2) + (-C + D) Add the last two simplified equations together to eliminate C & D: 0 = 2/a^2 + 2*D D = -1/a^2 Plug in to find B&C: 0 = 1/a^2 + B*(1 + a^2) + C - 1/a^2 0 = B*(1 + a^2) + C Only way that the a^2 term can be eliminated is if B=0. With equation 0=B+C, this also means C=0 Thus the Laplace transform becomes: 1/a^2/s^2 - 1/a^2 * 1/(s^2 + a^2) Multiply 2nd term by a/a to make a appear in the numerator: 1/a^2/s^2 - 1/a^3 * a/(s^2 + a^2) Inverse Laplace for our final answer: 1/a^2 - 1/a^3*sin(a*t)

[@dztekkz8494]

Sir please solve it using convolution Theorem

[@ManojKumarSahoo]

Please go through playlist

[@anamulislam4415]

sir last step se pehle (-1)kase aya

[@mralbaz716]

Sir please solve laplace inverse of 16÷(p^2+2p+5)^2

[@vijaysingh-lw7ct]

I want to get notes of inverse Laplace numericals

[@ankitprinja7748]

Great!!..I suggest you to remake this video in English. You will get much more views.By the way it was really helpful. Thankyou :)

[@amitverma-zm2us]

Hllo sir

[@JohnSmith-iu3fc]

I think your t should be moved to the denominator, not numerator. Is'nt it?

[@kausha.l404]

Sir ye Jo ap ne -1 tu smjh Aya Magar -1/2 kasie Aya mtlb sir 2 neche kasie Aya

[@bharatmishra7355]

Solving the d/ds will give you above eq.

[@kartikrajput878]

Sir thoda loudly bola kro

[@ajaykulkarni2386]

👌

[@mdnaseer4910]

Bhai teri awaaz hai ki nai??

[@muhammadumuhammad7850]

Same question fourier k liya kr dn plz

[@krezypandey7407]

Btao sir -t kaise aya

[@Abhishekyadav-xd2ib]

Why can't u simple break the term in two lart in starting

[@sagnikdingal5365]

-t×f(t) kaise aya?

[@aakashyadav2711]

The name of my mathematics teacher is also mks par wo mks ke nam pe dhabba he

[@krezypandey7407]

Btai sir -t kaise

[@umeshgautam3062]

Sir pls can u solve this question pls L ^-1 of s/s^4+4a^4

[@krezypandey7407]

-t

[@engineer2trader722]

kon #BDSK hai jo aisa aisa question ko dislike karta hai

[@bapunapradhan6131]

Like

[@ThanushaAsileti]

Video not clear sir

[@krezypandey7407]

-t btao

[@ps_online]

f(s)=1/s².(s-1/s+1) plizz

[@carultch]

Given: F(s) = 1/s^2 * (s - 1/s + 1) Distribute the s^2 to each term inside the parentheses: F(s) = 1/s - 1/s^3 + 1/s^2 These are all power rule terms, where L{t^n} = n!/s^(n + 1). Multiply where necessary to produce n! in the numerator, and divide by the same number: F(s) = 1/s - 1/2 * 2/s^3 + 1/s^2 F(s) = 0!/s^1 - 1/2* 2!/s^3 + 1!/s^2 Thus our inverse Laplace is: f(t) = 1 - 1/2*t^2 + t

[@jaskiratsingh6057]

Sir a second last step mein -1 and T Kahan Se Aaya

[@sandeeptiwari3655]

Jaskirat Singh ji -1 to pata nahi but "t" isliye aaya kyunki this is a property.........

[@krezypandey7407]

@@sandeeptiwari3655 😂

[@Ankityadav-pr5yf]

Sir mujhko clear ni hua hy

[@AnkitSharma-lr9do]

last me samaj nhi aaya question

[@chandansah1788]

sor

[@nishadkamble9330]

Bahot difficulty ho rahi hai samajhne me

[@jaheruddinshaik4311]

Ple rise your voice sound

[@manpreetsingh-ox1bo]

bilkul nhi smj aaiya sir last mai

[@eqrastudy4262]

Sir apka watsup no bolia na

[@amitverma-zm2us]

I amit from sonipat plz solve my 4question in mathmatics btech exam 2nd semester

[@anishachoudhury_]

Sir -1/2 kese aaya

[@premdeeptiwari3089]

minus half taken because when u differentiate 1/(s power 2 + a power 2) value will come minus 2s/(s power 2 + a power 2) whole power 2 , so minus half will cancel the minus 2 in numerator n dat will b equal to s/(s power 2 + a power 2) whole power 2

[@juliandrax4441]

Already there is all to the power '2' about ('s' power 2+ 'a' power 2), if we differentiate this, - 2 coming out from all to the power and again '2' come from 's' power 2 then, how - 1/2 is possible???

[@dineshmahawar4645]

Z transform kaha he sir 😭😭😭

[@ManojKumarSahoo]

Coming soon

[@dineshmahawar4645]

@@ManojKumarSahoo exam to aa gye lekin 15 se 15 se pahle kr dena upload Sir 🙏🙏

[@ManojKumarSahoo]

Ok brother

[@dineshmahawar4645]

@@ManojKumarSahoo Sir z transform upload kro jaldi

[@bitcoinwala1143]

Thank you Very much

[@vidyaprajapati9923]

Thank you sir

[@electrocode4095]

thanks sir

[@pkr7306]

Best expanded

[@pradipkumar-ll4oh]

👌

[@adarshsingh453]

Thank you sir