Answer to last example: y(t) = A*e^(s1*t) + B*e^(s2*t) + C*cos(ωt) + (D/ω)*sin(ωt) Thanks for the lectures. It's very helpful to shed light on many of the nuances I could not get from other similar lectures.
Yeah...lost audio at the last part of this video. For this particular example, Laplace transform is indeed much more complex than the usual method... But i think the best thing about Laplace transform is that it provides us with a reliable algorithm to solve for differential equation, i.e., the formalist approach instead of relying on intuition or great moment of Eureka.
Thank you for these outlines. Really helps me understand Laplace. While audio cut-out was unfortunate, I feel sorry for Prof. Strang. He seems to have known it cut out too, and looks crestfallen that he wasn't able to end with a concluding remark.
What confuses me is when Professor Strang takes a Laplace of a first or second derivative, he omits the y(0) and y'(0). What is his justification for doing that?
the kransfer function is G(s)=1/(s^2+Bs+c) only when initial conditions ( Y'(0)=0 , Y(0)=0 ) , is this really? if i am wrong , can you solve my doubt ? MIT OpenCourseWare
at 8:14 where did the y(0) terms go? i understand that the laplace transform of the derivative is s*Y(s) but there is an extra term that this guys omitted
I'm disappointed with the DE playlist. the calculus playlist is neat that's why I thought that this playlist will be also be of the same quality but it's not
The fastest trick for partial fractions, is Heaviside coverup. It works for the simple case of linear factors, and for the highest power of repeated linear terms. For irreducible quadratic factors, and for the remaining terms of repeated factors, it doesn't work. There are other tricks that work for those, such as letting s equal strategic values you haven't used yet, or taking the limit as s approaches infinity, and matching what remains. It helps to at least get the terms you can with Heaviside coverup, and simplify the rest of your work. The trick is, you cover-up the corresponding factor in the original expression, and then plug in strategic values of the variable, so that the term you covered up is equal to zero. As an example: (s + 1)/((s + 2)*(s + 3)) = A/(s + 2) + B/(s + 3) To find A, let s = -2, so that the (s+2) term equals zero. Cover up (s + 1) and evaluate the rest of the expression as s = - 2 to find A: A = (-2 + 1)/(-2 + 3) = -1 Likewise for B, at s = -3: B = (-3 + 1)/(-3 + 2) = +2 Thus, the partial fraction solution is: -1/(s + 2) + 2/(s + 3)
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I can binge-watch Prof. Gilbert Strang
Answer to last example: y(t) = A*e^(s1*t) + B*e^(s2*t) + C*cos(ωt) + (D/ω)*sin(ωt)
Thanks for the lectures. It's very helpful to shed light on many of the nuances I could not get from other similar lectures.
I really like the way MIT offers its lectures
The sound gets cut at 16:09. But this lecture really helped me understand more about the laplace transform, thank you!
Yeah...lost audio at the last part of this video. For this particular example, Laplace transform is indeed much more complex than the usual method... But i think the best thing about Laplace transform is that it provides us with a reliable algorithm to solve for differential equation, i.e., the formalist approach instead of relying on intuition or great moment of Eureka.
Excellent Teacher!!Sir the new generations are grateful!!!
Thank you for these outlines. Really helps me understand Laplace. While audio cut-out was unfortunate, I feel sorry for Prof. Strang. He seems to have known it cut out too, and looks crestfallen that he wasn't able to end with a concluding remark.
These mathematical tricks are awesome. I am constantly trying to learn them.
Strang added his comment that its zero initial condition : 12:20 hence the initial values are zero at LHS
Congratulations, your course is perfect.
Everything is clear and simple
Thx
I wish my professors had been as clear as these lectures! What a great style of teaching.
Love this guy!
we all do :')
What confuses me is when Professor Strang takes a Laplace of a first or second derivative, he omits the y(0) and y'(0). What is his justification for doing that?
the kransfer function is G(s)=1/(s^2+Bs+c) only when initial conditions ( Y'(0)=0 , Y(0)=0 ) , is this really? if i am wrong , can you solve my doubt ? MIT OpenCourseWare
see minute 12:20 , you have reason.The initial conditions are zero.
transfer*
This is great nice, shame about the sound cut-off at the end though
really really amazing
at 8:14 where did the y(0) terms go? i understand that the laplace transform of the derivative is s*Y(s) but there is an extra term that this guys omitted
He later mentioned that he is taking initial values of y and y' as zero
History doesn't repeat itself, but it mimes.
thank u sir
where are the proofs...
I'm disappointed with the DE playlist. the calculus playlist is neat that's why I thought that this playlist will be also be of the same quality but it's not
Chalculus...
I'd like to learn how he does his partial fractions so fast
The fastest trick for partial fractions, is Heaviside coverup. It works for the simple case of linear factors, and for the highest power of repeated linear terms. For irreducible quadratic factors, and for the remaining terms of repeated factors, it doesn't work. There are other tricks that work for those, such as letting s equal strategic values you haven't used yet, or taking the limit as s approaches infinity, and matching what remains. It helps to at least get the terms you can with Heaviside coverup, and simplify the rest of your work.
The trick is, you cover-up the corresponding factor in the original expression, and then plug in strategic values of the variable, so that the term you covered up is equal to zero.
As an example:
(s + 1)/((s + 2)*(s + 3)) = A/(s + 2) + B/(s + 3)
To find A, let s = -2, so that the (s+2) term equals zero. Cover up (s + 1) and evaluate the rest of the expression as s = - 2 to find A:
A = (-2 + 1)/(-2 + 3) = -1
Likewise for B, at s = -3:
B = (-3 + 1)/(-3 + 2) = +2
Thus, the partial fraction solution is:
-1/(s + 2) + 2/(s + 3)
@@carultch
Thanks!!
A sum of two logs is a product of a log.
wtf
Where should i start as a 14 year old?
in 18.01
It depends, what’s your math background?
calculus
I'd rather start in a football field :)
You learn transformations starting with watching Transformers.
During Re{} and Im{} treatment for cos(wt) as if he assumes 's' is real! This is shocking!
I totally agree to your analysis.
I like this guy much better than the MIT guys. the MIT guys waste too much time on theory, not on technique,