It's amazing how fast you switch from black marker to red marker when you're writing. The things you notice when you should be focusing on what's being written... :)
Hi, I do understand the math you're using however, could you tell me what is exactly a Laplace transform? in other words what are you exactly doing when you do this function?
Essentially, it's scanning the original function for a continuous composition of exponential decay functions, where s is the decay rate of the exponential decay, and the transformed function is the amplitude of each exponential decay term as a function of s. And it scans across the real domain of s, and the entire complex domain of s. Real values of s, are exponential functions, while imaginary values of s are sine waves. A complex combination is a product of an exponential and a sine wave. The Laplace transform scans for all of these component functions. A special case of the Laplace transform, is the Fourier transform, which is limited to the imaginary values of s. This corresponds to the steady state behavior after all transient exponential decays have completed.
i need to watch your video on integration by parts LOL, i tried to do it using the u du v dv method and got -2/s^3 instead... something went wrong with my calculation somewhere....
I could help you as to where you've gone wrong because I made a similar mistake. Make sure that when you are substituting infinite in for the t variables which gives you 0, that you properly write out the equation inclusive of 0. Do not ignore it! = 0 - (substitution of 0 into t variable) Then you'll quickly see that = 0 - (-2/s^3) = 2/s^3 Hope that helps! It's really easy to make the plus minus error especially if you discard the 0 and forget to include it in the equation to get the final answer.
LIATE/ILATE is good, and by good I mean it works around 75 to 85 percent of the time. But when it doesn't, it doesn't get you anywhere. So I wouldn't recommend using it.
Laplace transform of 2^t is 1/(s - ln(2)). Rewrite 2^t in terms of base e, so that it is e^(ln(2)*t). In general, the Laplace transform of e^(-k*t) is 1/(s + k). So when the coefficient on t is positive instead of negative, the addition turns into subtraction. The result is 1/(s - ln(2))
It's amazing how fast you switch from black marker to red marker when you're writing.
The things you notice when you should be focusing on what's being written... :)
your method for integration by parts is genius
fastest way to integrate by parts. Best method I've learned so far.
I love how you say "thank you" to all these handsome 0's :)
ohh!! what a fantastic explanation!!!
Thank you so much bor love from India , your video was so useful and helpful 😊
Thank you sir you made it very clear and I'll be taking my exams tomorrow
Godbless
Best of luck!
Gran ayuda y muy rápido que lo resolviste. Gracias bro
My black and red hero.
Thank you so much.
The capital n can be whatever you want,
ex: [lim,α->∞]([integral,0,α](e^(-st)*t^2 dt))
Thank you very much 🥰🥰
best one THANKK YOU
what happened to L'Hoptial's Rule?
Thanks a lot.
Hi, I do understand the math you're using however, could you tell me what is exactly a Laplace transform? in other words what are you exactly doing when you do this function?
en.wikipedia.org/wiki/Laplace_transform
Basically time domain to frequency domain.
Essentially, it's scanning the original function for a continuous composition of exponential decay functions, where s is the decay rate of the exponential decay, and the transformed function is the amplitude of each exponential decay term as a function of s. And it scans across the real domain of s, and the entire complex domain of s.
Real values of s, are exponential functions, while imaginary values of s are sine waves. A complex combination is a product of an exponential and a sine wave. The Laplace transform scans for all of these component functions.
A special case of the Laplace transform, is the Fourier transform, which is limited to the imaginary values of s. This corresponds to the steady state behavior after all transient exponential decays have completed.
thank you, you are awesome
Amazing thank you so much
i need to watch your video on integration by parts LOL, i tried to do it using the u du v dv method and got -2/s^3 instead... something went wrong with my calculation somewhere....
I could help you as to where you've gone wrong because I made a similar mistake. Make sure that when you are substituting infinite in for the t variables which gives you 0, that you properly write out the equation inclusive of 0. Do not ignore it!
= 0 - (substitution of 0 into t variable)
Then you'll quickly see that
= 0 - (-2/s^3) = 2/s^3
Hope that helps! It's really easy to make the plus minus error especially if you discard the 0 and forget to include it in the equation to get the final answer.
@@Lucylu723 Thank you! I appreciate your help
@@EricLeePiano No prob😊
perfect ........ thanks ....... please i want the Laplace application
Fucking legend. I would buy you a beer 🍻 if ever we met
Thanks! I appreciate it.
Thankyou
thanks bro!
Very helpful #💯
veri gud!!! tanku
i would be grateful if sir could upload a video in integration by parts by using (ILATE)rule in shortcut way..
LIATE/ILATE is good, and by good I mean it works around 75 to 85 percent of the time. But when it doesn't, it doesn't get you anywhere. So I wouldn't recommend using it.
Thanks sir
Why can't there be infinity-infinity in the first half of the integrated result??
you have to understand that infinity isn't a number. It's a concept or rather a representation.
Thank's :)
Marker switcher
i love you
Laplace transform of t cube
How to find the laplace transform of t^2 - 2t
I think you can break it in 2 integrals.(since you multiply both of them with e^(-st). Then you solve each of them.
The addition sign is used twice in this video. Twice.
Laplace transform of 4
what is ans of 2^t
Laplace transform of 2^t is 1/(s - ln(2)).
Rewrite 2^t in terms of base e, so that it is e^(ln(2)*t). In general, the Laplace transform of e^(-k*t) is 1/(s + k). So when the coefficient on t is positive instead of negative, the addition turns into subtraction. The result is 1/(s - ln(2))
Lalplace transform of 0
BolnA kya chahte ho
All i saw is + - + - diagonal diagonal diagonal. Fk that i'm outta here
That's the way integration by parts should be taught. I don't know why anyone bothers with the formula.
You plugged zero for N when it should be inf.
:) LOL t^n has a direct formula :P
Chinmay Vlogs
So?
Sometimes you need to prove the answer from the formula and i thank blackpenredpen that he doesnt think the way you do.
what is ans of 2^t