Response to Exponential Input

[@tigerwuli2760]

Still makes contributions to human beings even 82 years old. So moved. T T

[@freemanshell1913]

You can learn from the later video(solution for any input ) and realize the missing part (1.4a) might be dy/dt=ay+c.The particular solution is Yp=--c/a,the whole solution is y=[y(0)+c/a]e*at --c/a.

[@Gers217]

The fact that he takes the time to explain resonance is an absolute delight

[@ML-uu5ik]

Thank you so much for posting these! I'm reviewing diff eq over the summer and this series is very helpful. Dr. Strang explains things clearly and draws interesting connections I didn't know before.

[@zacharythatcher7328]

The very necessary intuition that this course skips is the fact that adding the equations for the full equation, and the equation for when the additional term is zero (with any arbitrary coefficient), provides a solution to the full equation (The ability to have any coefficient for the “null solution” is very important here, because it makes the initial value problem very easy to solve. It is possible because the null solution will never add anything to the additional term no matter what the constant coefficient is. It will always be contained to the original term by virtue of it spitting out zero as the additional term aka no additional term). If you only found the particular solution for the full equation then it would be very difficult/impossible to solve the initial value problem for any initial value. The “null” solution provides a very simple format for solving the initial value problem, and we know that the combination of it with the particular solution is a solution, so let’s just see if an equation in that format provides an easy/generalizable answer... and it does.

[@MrScattterbrain]

Prof. Strang speaks a lot about null and particular solution in his textbook. The video segments cover only a small part of the book (selected examples, I would say), and are pretty much useless without the text.

[@underfilho]

Now I can see, these videos are made for you to see while read the book, in this case looks like he's talking about non-homogenous linear equations and its relations with linear algebra, not just about methods to solve some differential equations. The null solutions yn are the solutions for the homogenous equations (like y'+y=0), the particular solution yp is the solution to the non homogenous equation (y'+y=q(t)), so the general solution is yn+yp. In the book this is explained, the videos are just extra content. It's like the differential equation now has a source term, that's q(t).

[@akshat_senpai]

Book 😅 name

[@tchappyha4034]

10:42 lim_{s -> a} [e^{s*t} - e^{a*t}] / [s - a] is the derivative of f(s) = e^{t*s} at s = a. So, it is t * e^{t*a}.

[@elamvaluthis7268]

Excellent derivation salutations to professor .

[@daisyjuarez6138]

I wish he was my grandpa. I could just imagine Christmas with him

[@postbodzapism]

The process of finding null solutions is equivalent to setting e^st =0, which means setting t to minus infinity. This is asymptotic analysis!

[@yuxiangfu4911]

you should read first 300 pages of the book (by C. Edwards (Author), David Penney (Author), David Calvis (Author)) to get something missing in the video.

[@alwaysbored47]

Thank you for this recommendation!

[@terryphi]

Am I missing something?

[@BenjaminEllenberger]

Watch the first 10 minutes of the first order differential equations part by Gilbert Strang, then it makes more sense: kzbin.info/www/bejne/aomThHp9p713n6s

[@RelevantDad]

@@BenjaminEllenberger Thanks!

[@yangliu4181]

@@BenjaminEllenberger The first 10-11 mins do solve my query, thank you.

[@atacoskun3597]

There are so many terms are definitions here comparing to last lecture. For example particular solution, null solution(I do not understand), homogenous vs... Do you know any other video lectures covered all definitions and topics of differential equations and linear algebra?

[@AhMedKhaled-om8ly]

yes , it is very advanced in the Subject , i struggle to understand

[@sharafmakk2936]

I do not recommend this series for beginners

[@bedwarri0r333]

Question: Towards the end, we apply l'Hospital's rule, which I am familiar with. Why is it that we can differentiate with respect to 's' and not 't'? The function Y(t) is a function of t. In that respect, wouldn't s be a constant?

[@tdchayes]

We are thinking about the result as 's' approaches 'a', so it is, in fact, the moving value. You could also think about 'a' approaching 's' instead - you get the same answer.

[@amiremami6259]

بی نظیر . خدا شما را حفظ کند

[@iwtwb8]

It seems like this lecture doesn't follow from the previous one. He implies there are things from previous lectures that haven't been covered. Are there missing lectures?

[@mitocw]

This video series corresponds to the textbook chapter numbers in Prof. Strang's textbook which should include the background materials needed. See the Differential Equations section of the OCW course site: ocw.mit.edu/RES-18-009F15.

[@konstantinosarsen581]

Well where can one find the missing content? I checked the website but i did not find something about the textbook, just the same lectures. Could you be more specific please? Thanks for your time

[@kasiviswanathboddeti9085]

@@dkirby6216 thanks

[@ahmedismail1018]

why is Yn=ce^at ? could you pls explain

[@maobesoya7335]

The homogeneous solution consider only: "y" elements, then you will get: dy/dt = a*y (1) No you apply separation of variables: dy/y = a*dt (2) integrating you get: Ln(y) = a*t + C (3) taking exponential on both sides e^(Ln(y)) = e^{at}*e^(c) renaming C= e^{c}, and operating Y = C*e^(at)

[@giangpham1946]

Can anyone explain for me why the general solution has the form y(p) + y(null)? Thanks!

[@Rayquesto]

Giang Pham Oh.... because it means the null space solution. Basically, whenever the variable side adds to zero, you get a null space solution (traditionally called the homogeneous solution). Typically, you will generate a system of differential equations and all the matix*vector(y)+matrix*vector(y')+matrix*vector(y'')+...+ matrix*vector(y(n)) is on one side and on the other side is a vector of constants b. If all entries in b go to zero, and a solution to y is found, then it is considered in the the nullspace. Now, yvp is solved by plugging in any arbitrary, but all possible orthogonal solutions with coefficients as unknown and once those are found, it becomes yvp (traditionally called the non-homogeneous solution). Note: If b were a sinusoidal input function (e.g. sin(t)) in which you have yvp(t), then the yvp(t) would be the steady-state solution to the differential equation. In any other case, I am not sure... so in other cases, to use a trial function with undetermined coefficients may be a lot more arbitrary than sins and cosines. The reason for this you can think is that the frequency of the output is the same as the input for the steady-state solution (again, emphasis should also be placed on the fact that each term is linear (e.g. no (y')^2, nor y^2, etc. terms in diff eqn or higher order).

[@DanielViteripixplanet]

Great question and great answer.

[@ozzyfromspace]

Another way to think of it is that 7 = 7 + 0. All solutions have that zero, because it doesn’t change anything (and as such, we call it null). So when Professor Strang solved dy/dt = a*y + e^(s*t), you could think of this as dy/dt - a*y = e^(s*t). If you solved the equation for the case where the right hand side is zero, your solution wouldn’t change anything, because said solution when plugged into the equation gives zero by definition. So you can add it to whatever the solution was with the e^(s*t) part - engineers call it the source, or driving, term. Rewatch the video and it should come together quite nicely :)

[@nehasehrawat1025]

@@ozzyfromspace thanks man!

[@ozzyfromspace]

@@nehasehrawat1025 definitely, glad it was helpful :) best wishes.

[@holydamn666]

Can someone help me to understand why a particular solution has the form of yp=e^st? I can see why the solution of y'=ay is y_0e^at but how can you see that a particular solution with the exponential input is yp=e^st

[@zacharythatcher7328]

The particular solution is a multiple of e^st. This is simply a good guess because if you take its derivative with respect to t, s comes down. Then, it is reasonable to believe that it can be a solution because the derivative is equal to itself plus an extra of itself, aka it is equal to some different multiple of itself. This, of course, depends on the exponent that is going to come down (s) and the coefficient that is in front of y (a), which is what we find when we solve for big Y (the multiplicity of the particular solution that is necessary to make the equation true). What is then done by solving for big Y, is finding the exact multiple in terms of the exponential coefficients that solves this equation.

[@guliyevshahriyar]

Thank you a ton, professor!

[@wanluz277]

Great illustration 👩🏼

[@jenniferlai8752]

it seems that from 1.2 to 1.4, 1.3 is missing? Did I miss anything here?

[@companymen42]

I wonder what happens if you take the limit as s -> a?

[@eeg9645]

Wooow!! Laplace Transform.

[@Andratos95]

Can someone explain to me why he talks about "sources" and "deposits", referring to e^(st)?

[@boxxer221]

I'm not completely sure but from what I gather the initial term is the [interest earned on your initial bank balance + your bank balance], you treat the interest as a continuous series of deposits of money which grows your bank balance, and then the interest that gets applied to the *additional money* you earned is the source term. So the initial term would be like say for example if you deposit £1000 with a 10% interest rate, then after a year your initial term would be £1000 + 10% = £1100, then the source term would be the interest earned on that additional £100 plus the interest on that interest, plus the interest on that interest etc, so the source term is always applying new interest to the additional money continuously. This continuous application of interest is what we call exponential growth and this is where the function comes from.

[@Arbiteroflife]

He talks about it more extensively in his other Diff. Eq. lecture series. The ones that have an actual class of students.

[@ricardoraymond9037]

God created the world and he created Gilbert Strang to complement his creation.

[@daisyjuarez6138]

Me when there is juicy drama 12:12

[@appleapple1763]

where could we see whole lectures?

[@UrViridescentLeaf]

Here kzbin.info/aero/PLUl4u3cNGP63oTpyxCMLKt_JmB0WtSZfG

[@nehasehrawat1025]

I didn't get the null equation at 4:05

[@muhammadfaizanalibutt4602]

How did he calculate null solution?

[@Amine-gz7gq]

the null (or homogeneous) solution is the solution to the homo equation : y' - a*y = 0 watch khan academy diff eq playlist if you want more explanations. homo eq is an equation where you only have y's, you don't have nasty stuff like exponentials or sines/cosines. why do we also use the null solution ? it's the solution when there's a zero on the right side equation and we always have a zero on the right side of the diff eq

[@flawfulpractice8381]

I guess s was a constant

[@flawfulpractice8381]

Then how could we apply L-hospital's RULE wrt s

[@physl2787]

Sir how from where you write particular solutiin

[@bigdave6952]

wow that was cool

[@vanvan22344]

where is lecture 1.3?

[@mitocw]

This video series is divided into eight parts corresponding to chapters of the textbook (1.2 refers to section 1.2 in the book). More information can be found on the course site: ocw.mit.edu/RES-18-009F15.

[@vanvan22344]

Thank you very much for the clarification

[@pubgplayer1720]

did he dye his hair? lol his hair is completely white in 2019

[@FernandoVinny]

so unorganized with the board