this interpretation is quite amazing the way it has been brought out is something so precise and inclusive that i am out of words.... well certainly prerequisite to understand this is more like that at least you had once met to all these equations earlier ... that integeral term and way it connects to fundamental theorem of integeration and way these deposits connecting to functions we met earlier is just awesome...
I understand this because i struggled with DE for so long because my own teacher introduced me into it very badly and out of the blue, then i watched tons of professor leonard, khan academy and 2blue1brown, then this is a wrap up from another perspective, more like a physicians perspective.
I am struggling with physics (I wanna do my phd in theoretical physics) and my math is absolutely horrible because the professors would just throw math and formulas in my face with no meaning behind it. My calculus and DE was extremely horrible because it made no sense, for me it was a tool to solve integrals with no meaning behind them. But now im re learning math slowly day by day and I see how beautiful and important math is and for explaining physics. Keep it up bruh! You are not alone in this fight.
You have to watch the fundamental theorem of calculus video in the beginning. The integral part is due to infinitesimal summation of interest compounding. The q(s) is the amount of money deposited. You get interest compounded from the amount you deposit each month. The beginning of the year you get the most interest compounded so e^(at) but at the middle or later of the year you less compounded, so e^(a(t-s)) a function of (s). This multiples your money deposited [e^(a(t-s))]*[q(s)] the infinitesimal summation is an integration. so Yp is integral of [e^(a(t-s))]*[q(s)] from 0 to t, ds. or in wolfram alpha Integrate[(e^(a(t-s)))(q(s)), {s, 0, t}]. The integration or summation of the compounded interest is the Yp. He wants to demonstrate Y'=aY + q(t). Take the e^(at) out, so that you have Integrate [ [e^(at)][e^(a)(-s)]*[q(s)] ] = [e^(at)] Integrate [ [e^(a)(-s)]*[q(s)] ], Then (d( [e^(at)] Integrate [ [e^(a)(-s)]*[q(s)] {s, 0, t} ] )/dt) is equal to (Y') . Gets by chain rule you get (d( [e^(at)] Integrate [ [e^(a)(-s)]*[q(s)] ] {s, 0, t} )/dt)= (a)[e^(at)] Integrate [ [e^(a)(-s)]*[q(s)] ] {s, 0, t}] + [e^(at)] d( Integrate [ [e^(a)(-s)]*[q(s)] {s, 0, t}] )/ds the SECOND integral by the fundamental theorem of calculus is equal to [e^(-at)]*q(t). Then: (d( [e^(at)] Integrate [ [e^(a)(-s)]*[q(s)] {s, 0, t} ] )/dt)= (a)[e^(at)] Integrate [ [e^(a)(-s)]*[q(s)] ] {s, 0, t}] + [e^(at)] [e^(-at)]*q(t) Then: (d( [e^(at)] Integrate [ [e^(a)(-s)]*[q(s)] {s, 0, t} ] )/dt)= (a)[e^(at)] Integrate [ [e^(a)(-s)]*[q(s)] ] {s, 0, t}] + (1)*q(t). Then: Sub Yp=[e^(at)] Integrate [ [e^(a)(-s)]*[q(s)] {s, 0, t} ] from the original: Integrate [ [e^(a)(t-s)]*[q(s)] {s, 0, t} ] with compound interest it's just that the constant t terms is out of the ds integral. You get: d(Yp)/dt = (a)(Yp) + 1*q(t) or (Yp)'=(a)(Yp)+q(t). Which is what he wants to demonstrate for the Yp.
at cca. 12:00 mr. Strang uses fundamental theorem of calculus, that is: derivative of integral of a function is an original function, but... the integral was with respect to s, while he took derivative with respect to t... is that generally valid operation?
Thank you, professor Strong. Your lecture is clear, logical and easy to understand. Sometimes i imagine what a beautiful thing it would be to take your watch these videos several years earlier.
this is like rocket science if one does not have a BSc Hons degree in engineering or mathematical sciences.I doubt if any layman can understand this at all. No way~ impossible.
You have to watch the fundamental theorem of calculus video in the beginning. The integral part is due to infinitesimal summation of interest compounding. The q(s) is the amount of money deposited. You get interest compounded from the amount you deposit each month. The beginning of the year you get the most interest compounded so e^(at) but at the middle or later of the year you less compounded, so e^(a(t-s)) a function of (s). This multiples your money deposited [e^(a(t-s))]*[q(s)] the infinitesimal summation is an integration. so Yp is integral of [e^(a(t-s))]*[q(s)] from 0 to t, ds. or in wolfram alpha Integrate[(e^(a(t-s)))(q(s)), {s, 0, t}]. The integration or summation of the compounded interest is the Yp. He wants to demonstrate Y'=aY + q(t). Take the e^(at) out, so that you have Integrate [ [e^(at)][e^(a)(-s)]*[q(s)] ] = [e^(at)] Integrate [ [e^(a)(-s)]*[q(s)] ], Then (d( [e^(at)] Integrate [ [e^(a)(-s)]*[q(s)] {s, 0, t} ] )/dt) is equal to (Y') . Gets by chain rule you get (d( [e^(at)] Integrate [ [e^(a)(-s)]*[q(s)] ] {s, 0, t} )/dt)= (a)[e^(at)] Integrate [ [e^(a)(-s)]*[q(s)] ] {s, 0, t}] + [e^(at)] d( Integrate [ [e^(a)(-s)]*[q(s)] {s, 0, t}] )/ds the SECOND integral by the fundamental theorem of calculus is equal to [e^(-at)]*q(t). Then: (d( [e^(at)] Integrate [ [e^(a)(-s)]*[q(s)] {s, 0, t} ] )/dt)= (a)[e^(at)] Integrate [ [e^(a)(-s)]*[q(s)] ] {s, 0, t}] + [e^(at)] [e^(-at)]*q(t) Then: (d( [e^(at)] Integrate [ [e^(a)(-s)]*[q(s)] {s, 0, t} ] )/dt)= (a)[e^(at)] Integrate [ [e^(a)(-s)]*[q(s)] ] {s, 0, t}] + (1)*q(t). Then: Sub Yp=[e^(at)] Integrate [ [e^(a)(-s)]*[q(s)] {s, 0, t} ] from the original: Integrate [ [e^(a)(t-s)]*[q(s)] {s, 0, t} ] with compound interest it's just that the constant t terms is out of the ds integral. You get: d(Yp)/dt = (a)(Yp) + 1*q(t) or (Yp)'=(a)(Yp)+q(t). Which is what he wants to demonstrate for the Yp.
Thanks MIT for this open course. I like Strang's teaching.
I got to understand intuitively the general solution of a linear differential equation thanks to this video. Thank you so much !
this interpretation is quite amazing the way it has been brought out is something so precise and inclusive that i am out of words....
well certainly prerequisite to understand this is more like that at least you had once met to all these equations earlier ...
that integeral term and way it connects to fundamental theorem of integeration and way these deposits connecting to functions we met earlier
is just awesome...
I understand this because i struggled with DE for so long because my own teacher introduced me into it very badly and out of the blue, then i watched tons of professor leonard, khan academy and 2blue1brown, then this is a wrap up from another perspective, more like a physicians perspective.
I am struggling with physics (I wanna do my phd in theoretical physics) and my math is absolutely horrible because the professors would just throw math and formulas in my face with no meaning behind it. My calculus and DE was extremely horrible because it made no sense, for me it was a tool to solve integrals with no meaning behind them. But now im re learning math slowly day by day and I see how beautiful and important math is and for explaining physics. Keep it up bruh! You are not alone in this fight.
2blue1brown😭🤣 I know it sucks
You have to watch the fundamental theorem of calculus video in the beginning. The integral part is due to infinitesimal summation of interest compounding. The q(s) is the amount of money deposited. You get interest compounded from the amount you deposit each month. The beginning of the year you get the most interest compounded so e^(at) but at the middle or later of the year you less compounded, so e^(a(t-s)) a function of (s).
This multiples your money deposited [e^(a(t-s))]*[q(s)] the infinitesimal summation is an integration. so Yp is integral of [e^(a(t-s))]*[q(s)] from 0 to t, ds. or in wolfram alpha Integrate[(e^(a(t-s)))(q(s)), {s, 0, t}]. The integration or summation of the compounded interest is the Yp.
He wants to demonstrate Y'=aY + q(t). Take the e^(at) out, so that you have Integrate [ [e^(at)][e^(a)(-s)]*[q(s)] ] = [e^(at)] Integrate [ [e^(a)(-s)]*[q(s)] ],
Then (d( [e^(at)] Integrate [ [e^(a)(-s)]*[q(s)] {s, 0, t} ] )/dt) is equal to (Y') . Gets by chain rule you get
(d( [e^(at)] Integrate [ [e^(a)(-s)]*[q(s)] ] {s, 0, t} )/dt)=
(a)[e^(at)] Integrate [ [e^(a)(-s)]*[q(s)] ] {s, 0, t}]
+
[e^(at)] d( Integrate [ [e^(a)(-s)]*[q(s)] {s, 0, t}] )/ds
the SECOND integral by the fundamental theorem of calculus is equal to [e^(-at)]*q(t).
Then:
(d( [e^(at)] Integrate [ [e^(a)(-s)]*[q(s)] {s, 0, t} ] )/dt)=
(a)[e^(at)] Integrate [ [e^(a)(-s)]*[q(s)] ] {s, 0, t}]
+
[e^(at)] [e^(-at)]*q(t)
Then:
(d( [e^(at)] Integrate [ [e^(a)(-s)]*[q(s)] {s, 0, t} ] )/dt)=
(a)[e^(at)] Integrate [ [e^(a)(-s)]*[q(s)] ] {s, 0, t}]
+
(1)*q(t).
Then:
Sub Yp=[e^(at)] Integrate [ [e^(a)(-s)]*[q(s)] {s, 0, t} ] from the original:
Integrate [ [e^(a)(t-s)]*[q(s)] {s, 0, t} ] with compound interest it's just that the constant t terms is out of the ds integral.
You get:
d(Yp)/dt =
(a)(Yp)
+
1*q(t)
or
(Yp)'=(a)(Yp)+q(t).
Which is what he wants to demonstrate for the Yp.
Thanks a lot for bringing attention on the first video, really helpful 👏
Excellent job, as always. BTW: ODE’s are wonderful partly because you could almost always find the solution (one way or another).
I am extremely grateful for these amazing posts. They are very helpful.
Where is the solution in complex form of the first order linear equations?
th-cam.com/video/Gp94Hph_-BU/w-d-xo.html
at cca. 12:00 mr. Strang uses fundamental theorem of calculus, that is: derivative of integral of a function is an original function, but... the integral was with respect to s, while he took derivative with respect to t... is that generally valid operation?
The integral will give you a function of t at the end, so taking the derivative with respect to t is the right way to do it
Thank you, professor Strong. Your lecture is clear, logical and easy to understand. Sometimes i imagine what a beautiful thing it would be to take your watch these videos several years earlier.
So much insight!
Splendidly done
where is the "first video" for q(s)=C? 5:00
I was making myself the same question.
Another excellence in its beauty.
Great job. So far so good 👌
this is like rocket science if one does not have a BSc Hons degree in engineering or mathematical sciences.I doubt if any layman can understand this at all. No way~ impossible.
You have to watch the fundamental theorem of calculus video in the beginning. The integral part is due to infinitesimal summation of interest compounding. The q(s) is the amount of money deposited. You get interest compounded from the amount you deposit each month. The beginning of the year you get the most interest compounded so e^(at) but at the middle or later of the year you less compounded, so e^(a(t-s)) a function of (s).
This multiples your money deposited [e^(a(t-s))]*[q(s)] the infinitesimal summation is an integration. so Yp is integral of [e^(a(t-s))]*[q(s)] from 0 to t, ds. or in wolfram alpha Integrate[(e^(a(t-s)))(q(s)), {s, 0, t}]. The integration or summation of the compounded interest is the Yp.
He wants to demonstrate Y'=aY + q(t). Take the e^(at) out, so that you have Integrate [ [e^(at)][e^(a)(-s)]*[q(s)] ] = [e^(at)] Integrate [ [e^(a)(-s)]*[q(s)] ],
Then (d( [e^(at)] Integrate [ [e^(a)(-s)]*[q(s)] {s, 0, t} ] )/dt) is equal to (Y') . Gets by chain rule you get
(d( [e^(at)] Integrate [ [e^(a)(-s)]*[q(s)] ] {s, 0, t} )/dt)=
(a)[e^(at)] Integrate [ [e^(a)(-s)]*[q(s)] ] {s, 0, t}]
+
[e^(at)] d( Integrate [ [e^(a)(-s)]*[q(s)] {s, 0, t}] )/ds
the SECOND integral by the fundamental theorem of calculus is equal to [e^(-at)]*q(t).
Then:
(d( [e^(at)] Integrate [ [e^(a)(-s)]*[q(s)] {s, 0, t} ] )/dt)=
(a)[e^(at)] Integrate [ [e^(a)(-s)]*[q(s)] ] {s, 0, t}]
+
[e^(at)] [e^(-at)]*q(t)
Then:
(d( [e^(at)] Integrate [ [e^(a)(-s)]*[q(s)] {s, 0, t} ] )/dt)=
(a)[e^(at)] Integrate [ [e^(a)(-s)]*[q(s)] ] {s, 0, t}]
+
(1)*q(t).
Then:
Sub Yp=[e^(at)] Integrate [ [e^(a)(-s)]*[q(s)] {s, 0, t} ] from the original:
Integrate [ [e^(a)(t-s)]*[q(s)] {s, 0, t} ] with compound interest it's just that the constant t terms is out of the ds integral.
You get:
d(Yp)/dt =
(a)(Yp)
+
1*q(t)
or
(Yp)'=(a)(Yp)+q(t).
Which is what he wants to demonstrate for the Yp.
@Mathe Mathe Same here. First semester of physics. Quite intuitive, when you understand the basics
I'm 16 years old and derived all of this before watching the video then checked I was right
@@Alexzuhnday Ok Einstein calm down.
It is quite easy if you have studied calculus.
Great lecture..
Where is the first video where the constant input case is discussed ?
Is this what you are looking for? th-cam.com/video/MJUjSKew4nQ/w-d-xo.html
@MIT openCourseWare No.. he is talking about lecture number 1.4a which is missing in the playlist and also not mentioned in the course website.
Here in this video he mentioned that first video was for solution for constant input that is missing here on TH-cam as well on Mit website
That's just fantastic ❤
This is actually lecture no 3
why?
Oh, and maybe I could include ........ 😀
Should you only use this formula if you have an initial value?
If y(0)=0, then you don’t need to write y(0).exp(at) term. The null solution will be 0.
there's my formula I'm hoping you will admire.
His explanations are not well structured and are quite confusing to follow through.