Just lovely. I can't stop being so amazed about how easy it is to spread information now a days. I can sit here in Denmark, with a hot cup of coffee, crying my eyes out over Laplace- and fourier-transforms due to an upcomming exam, and then some BRILLIANT australian saves the day. God bless the internet and Dr. Chris - thank you so very much.
you are so organised i felt like it is a mold and you are injecting the information into it, i have bee straggling for apx 6 Hrs and i could not understand , but with one hour with i got it .. thank you thank you thank you
Hi Chris, at 3:40 at the bottom of the page in the proof of L(e^-at g(t)) = G(s + a), I don't understand how you get from the 2nd last and last bits - the way I do it gives me G((s + a)^-1) Many thanks Euan
Technically speaking in your second example, a is equal to 1, not negative 1; the negative sign comes with the e^ -at. If a was actually equal to -1 then e^-at = e^ - (-1)t = e^t and the original f(t) would be (t+1)^2*e^-t. Seemingly trivial perhaps but actually important, especially when the second shifting theorem doesn't have that sign change
Allah bless you a million times over....so nice...the one dislike was possibly error. Statistically speaking a dislike is bound to happen. Hehe see what I did there.
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Just lovely.
I can't stop being so amazed about how easy it is to spread information now a days. I can sit here in Denmark, with a hot cup of coffee, crying my eyes out over Laplace- and fourier-transforms due to an upcomming exam, and then some BRILLIANT australian saves the day. God bless the internet and Dr. Chris - thank you so very much.
Saved me hours of frustration trying to dissect the textbook. Thank you Dr.
wooow, this is just amazing, been struggling for a while now , but finally i got it. a big thank from Norway Dr.
Wow, these videos might just be some of the best teaching material I've been lucky enough to witness. Keep it up!
Your a life saver ...That just saved me hours of trying to figure it out with my text book ..Thank you
you are so organised i felt like it is a mold and you are injecting the information into it, i have bee straggling for apx 6 Hrs and i could not understand , but with one hour with i got it .. thank you thank you thank you
Thank you for the great feedback!
Isn't this professor totally awesome?
For having bad notation, being stressed out? Not really ,no.
***** Bad notation? Stressed out? Please explain.
Dr Chris Tisdell some people are born negative keep up the good work man.
Excellent lectures.Dr.Chris made it very easy
Cheers Chris, really appreciate these videos don't know what I'd have done!
Thank you Sir for this and other great series of yours in the field of mathematics! Greetings from Trondheim, Norway!
respect from Jamaica DR
Thanks for the great feedback.
A big thank you from Canada
Thank you so much sir....ur lecture helped me so much fr my math.......
Thank you so much sir...this thing really really helped...!
thanks so much Dr. Chris
Dr may God bless you .
that is just "perfect" , thanks very much
You are welcome!
hi Dr : can we solve laplace equation by lablace transformation ? thanks
thanks Buddy it helps me a lot
Respect from India
big thank u from south africa,that was awesome
7
Hi Chris,
at 3:40 at the bottom of the page in the proof of L(e^-at g(t)) = G(s + a), I don't understand how you get from the 2nd last and last bits - the way I do it gives me G((s + a)^-1)
Many thanks
Euan
THANK YOU!!!! TY! TY! TY! You are awesome!
It is my pleasure!
thx you very big knowledge. thailand
Technically speaking in your second example, a is equal to 1, not negative 1; the negative sign comes with the e^ -at. If a was actually equal to -1 then e^-at = e^ - (-1)t = e^t and the original f(t) would be (t+1)^2*e^-t.
Seemingly trivial perhaps but actually important, especially when the second shifting theorem doesn't have that sign change
sir one more question (t+1) ka whole square e power t ka first shifting property
What it is within a function i.e sin(te^(-2t))
Thank u
Sir u hve not written 2! factorial in calc. Remainig procedure is awesome.
thanks:)
thanks a lottttt sir :-)
could you please solve L[t square sinat] would be really, helpful thanks.
+Hamza Kirax
L{sinat} = a/(s^2+a^2)
and L{t^2}=2!/s^(2+1)
hence by first shifting......
L{t^2.sinat}= 2a/(s^2+a^2)^3
Ohh thank you sirr :)
4:18 example
Allah bless you a million times over....so nice...the one dislike was possibly error. Statistically speaking a dislike is bound to happen. Hehe see what I did there.
It's my pleasure!
sir one more question (t+1) ka whole square e power t ka first shifting property