Inverse Laplace transform with unit step function, sect7.6#13

[@jonelquitiquit5529]

This is easier than i thought! Thanks man. You save me from my exams.

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Are you actually saved?

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[@prettylittlething6949]

I'm not sure whether i would excel in my assignment, but this surely helps alot. Thank you sir!

[@JuliaPaler]

This has never been so clear.... your videos are amazing. Please continue making them!!!

[@fhffhff]

e^-2t*t^1-3e^-2t*t^4

[@ibrahimudida3452]

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You're welcome!

[@philcooper279]

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And you are also motivating. Thank you so much!

[@richi3141]

First I thought you forgot to add the Heaviside functions, we use sigma instead of "u" at our university :D Your Videos are very helpfull and well explained! Greetings from Germany!

[@blackpenredpen]

richi hill you're welcome!! It's my pleasure to help!!

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Aleksja Braka you are so welcome!!! I am glad to have you watch my videos :)

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[@blackpenredpen]

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[@michaeldinardi3237]

i love you so much this is the exact problem i am stuck on

[@rohitreddy1340]

very nice explanation

[@johnmarcano952]

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[@swatiagarwal5097]

symmetric function of roots of cubic and biquadratic equation

[@shissmissshrishti]

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[@dnuyc]

THANK YOU

[@muhammadsaqib2400]

Please suggest a book. Thanks

[@leightonater98]

Pretty sure he is using Fundamentals Of Differential Equations by Nagle Saff and Snider. 8th and 9th edition have basically the same questions. But the sections and questions match up with my book which is the 9th edition so far. Actually just made this realization and its very exciting for me since this is the book my professor uses!!

[@lj54]

I agree with leightonater98. Pretty sure he's using the textbook: "Fundamentals of Differential Equations" by Nagle Saff and Snider

[@mathscience15]

Great

[@ajayvaghasiya7009]

thank yu so much sir

[@Egghead991]

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[@sasiraja6258]

The inverse laplace transform of 1/s(s^2+a^2) is

[@carultch]

Given: £{f(t)} = 1/(s*(s^2 + a^2)) Do a partial fraction decomposition. 1/(s*(s^2 + a^2)) = A/s + (B*s + C)/(s^2 + a^2) Heaviside coverup tells us that we can let s=0, cover up the s-factor, and plug in s=0 to find A: A = 1/(0^2 + a^2) A = 1/a^2 Thus: 1/(s*(s^2 + a^2)) = (1/a^2)/s + (B*s + C)/(s^2 + a^2) Multiply out the sum of fractions: (1/a^2)*s^2 + 1 + B*s^2 + C*s = 1 Equate coefficients: 0 = 1/a^2 + B 0 = C Solve for B: B = -1/a^2. Reconstruct: 1/(s*(s^2 + a^2)) = (1/a^2)/s - (1/a^2)*s/(s^2 + a^2) The first term resembles the Laplace of f(t) = 1, which is simply 1/s, but with a coefficient of 1/a^2. This means the first term in our inverse Laplace is 1/a^2. The second term resembles the Laplace of f(t) = cos(a*t), except with a leading constant of -1/a^2. This means the 2nd term in our inverse Laplace is -1/a^2 * cos(a*t) Conclusion: £^-1 {1/(s*(s^2 + a^2))} = 1/a^2 * (1 - cos(a*t))

[@sosoalsabahi3784]

Cooooool

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Ayyy thanks

[@СергейВладимирович-к2з]

How to solve inverse Laplace transfom of exp(-pi*sqrt(s)) ?

[@Acoustic_Mutale]

MAT 3110 with Mr. Trevor anyone?

[@sasiraja6258]

The inverse laplace transform of 1/s(s^2+a^2) is