Started the video, paused, went to my marker board and solved it! Then skipped to the end and I saw that I got it right. First time I have solved one of these on my own. Your help has been HUGE! Thank you!
I want to comment that you can use residue method to get A/s. To do so, we will consider P(s) = (s^3 + 2s^2 + 1)/(s^2 + 4). Then, we can differentiate P(s) but we have to divide it by n! where n is the number of times differentiation can occur. (In this case, n= 1) because s is only repeated once. P'(s) = [(3s^2 + 2s)(s^2 + 4) - 2s(s^3 + 2s^2 + 1)]/(s^2 + 4)^2 From the previous cover-up: s = 0 so we have P'(0). Then, you can plug 0 into your function. After applying differentiation. You'll be surprised you'll get the same answer for A.
You can use cover-up to solve for C and D. Just make s=2i and substitute in. You get 2i+(-7)/(-4)=2i*C + D, therefore D = 7/4 and 2i=2i*C, thus C = 1. BAM!
Given: (s^3 + 2*s^2 + 1)/(s^2*(s^2 + 4)) I like to set up the terms for Heaviside coverup first, which in this case is A: A/s^2 + B/s + (C*s + D)/(s^2 + 4) At s=0, cover up s^2, and find A: A = (0+0+1)/(0^2 + 4) = 1/4 Reconstruct: (s^3 + 2*s^2 + 1)/(s^2*(s^2 + 4)) = 1/4/s^2 + B/s + (C*s + D)/(s^2 + 4) Multiply by s, to partially clear the fraction: (s^3 + 2*s^2 + 1)/(s*(s^2 + 4)) = 1/4/s + B + (C*s^2 + D*s)/(s^2 + 4) Take the limit as s goes to infinity, to set up our first equation: 1 = 0 + B + C B = 1 - C Pick two values of s we haven't used yet, to create two more equations. I'll choose s=1 & s = -1 s=1: (1^3 + 2*1^2 + 1)/(1^2*(1^2 + 4)) = 1/4 + B + (C + D)/(1 + 4) 4/5 = 1/4 + B + (C + D)/5 20*B + 4*C + 4*D = 11 s = -1: ((-1)^3 + 2*(-1)^2 + 1)/((-1)^2*((-1)^2 + 4)) = 1/4/(-1)^2 + B/(-1) + (C*(-1) + D)/((-1)^2 + 4) 2/5 = 1/4 + -B + (-C + D)/5 8 = 5 + -20*B + 4*(-C + D) -20*B - 4*C + 4*D = 3 Add equations to cancel B & C terms and solve for D: 8*D = 14 D = 7/4 Use original equations to solve for B&C: 20*B + 4*C + 4*7/4 = 11 5*(1 - C) + C = 1 -4*C + 5 = 1 C = 1 B = 1-1 = 0 Result: 1/4/s^2 + (s + 7/4)/(s^2 + 4)
Your tutorials are the best just that I'd like to comment on this and a previous video about this same IVP that you should note the way you write your (Y and y). It conflicts as to know which is the laplace and the inverse laplace, Just to help amateur like me. Thank you for your best explanation.
I dont really know why professors in my uni just does not go into this much depth into explaining a single example. The way he teaches me reminds me of my high school teacher who used to make complex problems easily digestible.
Seriously one of the best teachers I've ever encountered both on youtube and real life. I salute you! Thank you for explaining every detail of your solving and making sure we understand every step you take
For linear terms, you only need a constant for the numerator. For irreducible quadratic terms, you need to set up a linear term for the numerator. If you had an irreducible cubic, you'd use a quadratic term for its numerator. It wouldn't help you very much for either integration or Laplace transforms, but that's what you'd do in concept. In general, the polynomial on the top, is one degree less than the polynomial on the bottom. Sometimes, the linear factor turns out to just be a constant, other times, the linear factor turns out to only be the term with the variable
Great job man, really good explanation. I think whats so hard to understand is how the denominator is split up into A, B etc. I've taken earlier courses that covered it, but not in the same extend that these exercises need.
@@blackpenredpen Actually, I totally confused myself while I was doing the problem along with you. I missed an "s." It made perfect sense! Sorry about the inconvenience!
Literally the partial fractions is so unnecessary and it takes up over half the video, without taking the common denominator and keeping it as three seprate fractions we coud have taken the inverse laplace much easier with a VERY easy convolution to solve at the end
yes but in general you want to avoid convolutions. the point of the laplace transform is to solve the problem without actually doing calculus, so if you can find partial fractions then you should do that instead
Pointing out the seemingly obvious, but you forgot to multiply by 2 inside the cosine inverse LaPlace and the 1/2 outside, so it should be 1/2* cos(2t)
Started the video, paused, went to my marker board and solved it! Then skipped to the end and I saw that I got it right. First time I have solved one of these on my own. Your help has been HUGE! Thank you!
I am very glad to hear this!!!!
Spent the last week panicking about Laplace transforms - all of that anxiety vanished after watching this video! Thank you
Glad to hear!!
I want to comment that you can use residue method to get A/s.
To do so, we will consider P(s) = (s^3 + 2s^2 + 1)/(s^2 + 4).
Then, we can differentiate P(s) but we have to divide it by n! where n is the number of times differentiation can occur. (In this case, n= 1) because s is only repeated once.
P'(s) = [(3s^2 + 2s)(s^2 + 4) - 2s(s^3 + 2s^2 + 1)]/(s^2 + 4)^2
From the previous cover-up: s = 0 so we have P'(0).
Then, you can plug 0 into your function. After applying differentiation. You'll be surprised you'll get the same answer for A.
You just taught me more than my professor taught me in a whole semester❤
Yes, I just said "isn't it?" just like Y O U ! ! !
Yay!!!!!!!
You're a good tutor! Im about to comment because i got confused in the A/s +B/s2 ... but you discussed it well ahhaha slow clap*
Thank you so much dear Teacher 💖
You can use cover-up to solve for C and D. Just make s=2i and substitute in. You get 2i+(-7)/(-4)=2i*C + D, therefore D = 7/4 and 2i=2i*C, thus C = 1. BAM!
学到了啊 待定系数还能这么拆 很棒的细节讲解
Dude You are the best !!!!!! I owe You 20 Marks!!!
Thank you!!!!!!!
I distinguish my s and my 5 using cursive handwriting
i am a student on uni. at south korea.
thx your class. it a big help for me.
Just for the sake of trying the convolution formula i solved the partial fractions s/(4+s^2) + 2/(4+s^2) + 1/[s^2(4+s^4)]. Nice!
how we can get D=7/4
like how 2 become 8/4 - 1/4 = 7/4
just how that 2 become 8/4
sorry if my question is so complicated .
Given:
(s^3 + 2*s^2 + 1)/(s^2*(s^2 + 4))
I like to set up the terms for Heaviside coverup first, which in this case is A:
A/s^2 + B/s + (C*s + D)/(s^2 + 4)
At s=0, cover up s^2, and find A:
A = (0+0+1)/(0^2 + 4) = 1/4
Reconstruct:
(s^3 + 2*s^2 + 1)/(s^2*(s^2 + 4)) = 1/4/s^2 + B/s + (C*s + D)/(s^2 + 4)
Multiply by s, to partially clear the fraction:
(s^3 + 2*s^2 + 1)/(s*(s^2 + 4)) = 1/4/s + B + (C*s^2 + D*s)/(s^2 + 4)
Take the limit as s goes to infinity, to set up our first equation:
1 = 0 + B + C
B = 1 - C
Pick two values of s we haven't used yet, to create two more equations. I'll choose s=1 & s = -1
s=1:
(1^3 + 2*1^2 + 1)/(1^2*(1^2 + 4)) = 1/4 + B + (C + D)/(1 + 4)
4/5 = 1/4 + B + (C + D)/5
20*B + 4*C + 4*D = 11
s = -1:
((-1)^3 + 2*(-1)^2 + 1)/((-1)^2*((-1)^2 + 4)) = 1/4/(-1)^2 + B/(-1) + (C*(-1) + D)/((-1)^2 + 4)
2/5 = 1/4 + -B + (-C + D)/5
8 = 5 + -20*B + 4*(-C + D)
-20*B - 4*C + 4*D = 3
Add equations to cancel B & C terms and solve for D:
8*D = 14
D = 7/4
Use original equations to solve for B&C:
20*B + 4*C + 4*7/4 = 11
5*(1 - C) + C = 1
-4*C + 5 = 1
C = 1
B = 1-1 = 0
Result:
1/4/s^2 + (s + 7/4)/(s^2 + 4)
Your tutorials are the best just that I'd like to comment on this and a previous video about this same IVP that you should note the way you write your (Y and y). It conflicts as to know which is the laplace and the inverse laplace, Just to help amateur like me. Thank you for your best explanation.
Bro thanks so much for all youve taught me, in this case wouldn't it have been easier to solve using method of unconfirmed...
not difficult,
but very easy to make a mistake
His using 2 markers one hand
I dont really know why professors in my uni just does not go into this much depth into explaining a single example. The way he teaches me reminds me of my high school teacher who used to make complex problems easily digestible.
Seriously one of the best teachers I've ever encountered both on youtube and real life. I salute you! Thank you for explaining every detail of your solving and making sure we understand every step you take
Sorry man but I had to....
Pause the video to comment and give you a thumbs up! So glad I subscribed, great work!! Thank you!!
This vid helped me more then books
this guy saved my life
may i know why so i have to add CS + D and your peevious dont need?
For linear terms, you only need a constant for the numerator.
For irreducible quadratic terms, you need to set up a linear term for the numerator.
If you had an irreducible cubic, you'd use a quadratic term for its numerator. It wouldn't help you very much for either integration or Laplace transforms, but that's what you'd do in concept.
In general, the polynomial on the top, is one degree less than the polynomial on the bottom. Sometimes, the linear factor turns out to just be a constant, other times, the linear factor turns out to only be the term with the variable
Thanks a ton, now I feel very motivated
that pen swish doh 15:10. Swag
If that s³+2s²+1 was s³+2s²+s, you would do: s³+2s²+s=s(s²+2s+1)=s(s+1)²
I can't find your Laplace playlist.
Absolutely perfect job. Thanks!
is there a playlist for this kind of problem's?
I love youuuuu
Best way to solve DE is with Laplace Tf
agreed, unless u dont have initial conditions then my favorite way is variation of parameters
Thank you
Me gustaría entenderte en el idioma pero observando tu procedimiento me a ayudado, gracias
Math is beautiful
Lots of love for you sir.. 😍😍😍
From Bangladesh..
Your method was super easy to get the problem..
Watching this during my Calc 4 exam, thank you very much my dude
Great work! Technically shouldn’t all of this be multiplied times the unit step function, which would be the same as saying “for t >= 0?”
thank you
Aha! You want to look the 4 as 2², isn't it?
your video makes concept crystal clear
I don't sub often, I am about to graduate this year and you earned it!
your voice is asmr to my ears
Great job man, really good explanation. I think whats so hard to understand is how the denominator is split up into A, B etc. I've taken earlier courses that covered it, but not in the same extend that these exercises need.
look up "partial fraction decomposition"
thank you so much for explaining where the partial fraction decomposition comes from!
I found "A" to be 1/2 no zero
I don't understand at 6:45
How when 1/4 we put in A
woaahh i never knew laplace is this easy..thank you sir, tutorial appreciated!
Is that a microphone
Weird thing. My teacher use this equation in our assignment
Did you write your answer in black pen and red pen?
this deserves more likes than those useless travel vlogs
very kind smile
Love this!
wow. thank you
This is so tough
The best
Thank you so much!😭😭💕💕
The way you do partial fractions doesn't make a bit of sense...
Which part confused you?
@@blackpenredpen Actually, I totally confused myself while I was doing the problem along with you. I missed an "s." It made perfect sense! Sorry about the inconvenience!
Joshua Mitchell I see. 😎
You've helped me a lot thanks
ليه حطيت T=1/s^2 ???
why ?
Bedankt maat!
What is the laplace of sint
1/(s^2 + 1)
Cosine has the s up top, sine has the constant.
Thank you
Very good
As long as you are getting him
Worst way to solve a linear ODE. How about a non-linear ODE example?
Hi Rob!
lol dude - its about learning the method
dude its just an example to get the hang of laplace transforms chill out dickwad
hm, so you don`t use the +1...
Good observation, did you forget to use the +1?
hahahahaha temos um ancap por aqui.
5:00 20 seconds does not equal 50 seconds unless we are moving fast away from each other and relativistic effects take over.
Literally the partial fractions is so unnecessary and it takes up over half the video, without taking the common denominator and keeping it as three seprate fractions we coud have taken the inverse laplace much easier with a VERY easy convolution to solve at the end
yes but in general you want to avoid convolutions. the point of the laplace transform is to solve the problem without actually doing calculus, so if you can find partial fractions then you should do that instead
Pointing out the seemingly obvious, but you forgot to multiply by 2 inside the cosine inverse LaPlace and the 1/2 outside, so it should be 1/2* cos(2t)
Steven Tucker u only have to do that for sine. Cos is ok
But… no.
His English is so bad
better than yours