Inverse Laplace Transform of s/(s + 1)^2

[@alejandrosalinaslopez4411]

I don't understand much english but i think math is an universal lenguaje. Thanks good man.

[@TheMathSorcerer]

You are welcome!

[@teferasinara1768]

Your videos are helping me for upcoming exams. But I want to say one thing only, l will expect a lot better videos with you, thank you so very much

[@pruthveshmandle4978]

Thank You, Sir Issac Newton.

[@ShanBojack]

😂

[@MdMahafuzurRahman-y3o]

That is really helpful.. Thanks a lot.

[@ayushbansal7061]

Thank you so much 💓

[@crispindebolivia5539]

Me ayudó mucho, gracias 👍

[@TheMathSorcerer]

👍

[@parrot...3055]

Which country u belong.?? Is India???? (

[@TheMathSorcerer]

No I'm in the US

[@MohammadAfzal57]

Thank you 😊

[@bora3649]

it helped me a lot.

[@mohanramg1565]

Thank you

[@musachohan878]

Awesome teacher

[@ricardogarzacorona630]

thank you very much, you're awesome

[@anuraagvaddi5272]

Thank you so much!!

[@TheMathSorcerer]

You're welcome!

[@dikshagoyal5526]

Thanks a lot

[@Jason-ui9ue]

thank!! the video is useful for me

[@TheMathSorcerer]

Awesome !

[@young5668-f8s]

감사합니다 감사합니다

[@ahalyas6044]

Can u please tell the answer of the question u gave in the last to try ??

[@evilbaymax5667]

Can you help me solve the exact problem but with Heaviside's expansion theorem on partial fractions method, please?!

[@evilbaymax5667]

I've no idea how to find the same result as yours in video🙈

[@carultch]

@@evilbaymax5667 An example where Heaviside's method finds all coefficients, is the following: (3*s^2 + 2*s + 3)/(s*(s + 1)*(s + 3)) Generally, Heaviside coverup helps you for simple linear factors, and stand-alone variables as factors. It helps you find the highest ordered fraction of repeated roots as well, but then you need another method to find the remaining coefficients. I prefer plugging in strategic s-values to find the other coefficients. Set up partial fractions: A/s + B/(s + 1) + C/(s + 3) To find A, B, and C, cover up the corresponding fraction in the denominator (which I'll mark with #), and find the s-value that makes what you covered up, equal to zero. Using s= 0: A = (3*0^2 + 2*0 + 3)/(#*(0 + 1)*(0 + 3)) = 3/3 = A = 1 Using s = -1: B = (3*(-1)^2 + 2*-1 + 3)/(-1*(###)*(-1 + 3)) = (3 - 2 + 3)/(-1) = B = -4 Using s = -3: C = (3*(-3)^2 + 2*(-3) + 3)/(-3*(-3 + 1)*(###)) = (27 - 6 + 3)/(12) = C = 2 Thus the partial fraction result is: 1/s - 4/(s + 1) + 2/(s - 3) Invert the Laplace transform, to get the solution: 1 - 4*e^(-t) + 2*e^(-3*t)

[@LolLipop20898]

thank you sir

[@swezaldias3198]

Thank you sir

[@TheMathSorcerer]

You are welcome!

[@aniketgope9293]

Thankyou so much sir 💖

[@periswambui1553]

How about laplace inverse of 1/S^2+S+1

[@TheMathSorcerer]

Complete the square 😎I might have this video will look tomorrow but yeah, complete the square and use the first translation theorem and you got it,👍

[@periswambui1553]

@@TheMathSorcerer okay thanks though i would appreciate if you did a video on that

[@Lightofthatstar]

Why couldn’t we write cos?

[@albertoalmeida503]

Obrigado Achei que teria que usar frações parciais

[@DeepanshuSharma-tc9lr]

Good in this question i got an important formula that was not in my book. Thanks i am from india and you??

[@TheMathSorcerer]

Very cool! I live in the US

[@luson_18]

What if we have s² instead of s in the numerator?

[@carultch]

Given: s^2/(s + 1)^2 Add zero in a fancy way, to form a term we can reduce to 1: (s^2 + 2*s + 1 - 2*s - 1)/(s + 1)^2 Gather the perfect square: ((s + 1)^2 - 2*s - 1)/(s + 1)^2 Split the fraction and simplify the first term to 1: (s + 1)^2/(s + 1)^2 - (2*s + 1)/(s + 1)^2 1 - (2*s + 1)/(s + 1)^2 Again, add zero in a fancy way, to form a term we can cancel: (2*s + 1 + 2 - 2)/(s + 1)^2 = 2*(s + 1)/(s+1)^2 - 1/(s + 1)^2 = 2/(s + 1) - 1/(s + 1)^2 Reconstruct: 1 - 2/(s + 1) + 1/(s + 1)^2 Inverse Laplace of first two terms: L^-1{1} = delta(t) L^-1{-2/(s + 1)} = -2*e^(-t) For the third term, let G(s) = 1/(s + 1), which means the corresponding g(t) = e^(-t). Using the s-derivative property, we can convert G(s) into the function we have: L^-1{-d/ds G(s)} = t*g(t) -d/ds G(s) = 1/(s + 1)^2 Thus: L^-1{1/(s + 1)^2} = t*e^(-t) And we can construct the solution: L^-1{s^2/(s + 1)^2} = delta(t) - 2*e^(-t) + t*e^(-t)

[@الاستاذقاسمالكرعاوي]

❤❤❤❤❤❤

[@domeniccosio1859]

🐐

[@KangarooJack420]

this guy kinda looks like jeff bezos ngl

[@fodderking]

THANK YOU BASED GOD

[@TheMathSorcerer]

❤️

[@zakesrangoaga645]

How about s/(s^2 + 1)^3

[@carultch]

Given s/(s^2 + 1)^3. Assume a solution that is the following linear combination: A*cos(t) + B*sin(t) + C*t*cos(t) + D*t*sin(t) + E*t^2*cos(t) + F*t^2*sin(t) Take the Laplace transforms of each of the above, using the s-derivative property. L{cos(t)} = s/(s^2 + 1) L{sin(t)} = 1/(s^2 + 1) L{t*cos(t)} = -d/ds s/(s^2 + 1) = (s^2 - 1)/(s^2 + 1)^2 L{t*sin(t)} = -d/ds 1/(s^2 + 1) = 2*s/(s^2 + 1)^2 L{t^2*cos(t)} = -d/ds (s^2 - 1)/(s^2 + 1)^2 = (2*s*(s^2 - 3))/(s^2 + 1)^3 L{t^2*sin(t)} = -d/ds 2*s/(s^2 + 1)^2 = (6*s^2 - 2)/(s^2 + 1)^3 Construct an arbitrary linear combination, and equate to the original: s/(s^2 + 1)^3 = A*s/(s^2 + 1) + B/(s^2 + 1) + C*(s^2 - 1)/(s^2 + 1)^2 + D*2*s/(s^2 + 1)^2 + E*(2*s*(s^2 - 3))/(s^2 + 1)^3 + F*(6*s^2 - 2)/(s^2 + 1)^3 Clear denominators, expand, and gather: s = A*s*(s^2 + 1)^2 + B*(s^2 + 1)^2 + C*(s^2 - 1)*(s^2 + 1) + D*2*s*(s^2 + 1) + E*(2*s*(s^2 - 3)) + F*(6*s^2 - 2) s = A*s^5 + 2*A*s^3 + A*s + B*s^4 + 2*B*s^2 + B + C*s^4 - C + 2*D*s^3 + 2*D*s + 2*E*s^3 - 6*E*s + 6*F*s^2 - 2*F s = A*s^5 + (B + C)*s^4 + (2*A + 2*D + 2*E)*s^3 + (2*B + 6*F)*s^2 + (2*D - 6*E)*s + B - C - 2*F Equate like coefficients: A = 0 B + C = 0 2*A + 2*D + 2*E = 0 2*B + 6*F = 0 2*D - 6*E = 1 B - C - 2*F = 0 Solve system of equations for the variables: A = 0, B = 0, C = 0, D = 1/8, E = -1/8, F = 0 Thus, the solution is: 1/8*t*sin(t) - 1/8*t*cos(t)^2

[@mrbk1120]

The answer is te^-t

[@Knightfire66]

u explained it waaay to complicated. do +1-1 trick s+q/(s+q)^2 - 1/(s+1)^2 now s+q and ()^2 of (s+q)^2 do cancel each other. now we have 1(s+q) - 1/(s+1)^2 and using the standard formula for e^(at) we get e^-t - t*e^-t

[@faye2266]

I am dumb

[@pokurideepthi]

Thank you so much 😊

[@TonyLee34110]

Thank you