@@JDMaxton1999 Basically, it's not so much of an assumption, but more like an "educated guess". Whomever figured this out realized that making y1v1' + y2v2' = 0 would cut our problem in half, and has a nice form to plug into the DE. You can use any random number as a guess, but what you end up with is some nasty or potentially unsolvable DE problems. So this is a nice way to find a general form. Like BPRP was saying in his video, if we kept the first derivative as is, then when we plug it back into our original DE, we will be running into a second order DE within a second order DE. Just nasty, nasty stuff. Remember, c1y1 + c2y2 is A fundamental set, not THE fundamental set. There is potentially more than one answer for these problems based upon what initial conditions are given to either the constants or what f(t) is. This is just THE fundamental set based upon the given conditions. I hoped this helps. I'm also not a fan of "hand wave, black box magic" with math, but sometimes it's better to trust the system then learn later on why. I find this to be the "if your quadratic is becoming unsolvable, see if you can graph it and get an answer that way" equivalent in DE.
The explanation is given by the *second* answer to the math stackexchange question titled "Justifying an Assumption Made While Deriving the Method of Variation of Parameters". It explains the *simple* reason why 'osculating functions' will always satisfy this constraint. I have no idea why this step is never justified, now that I know the answer is so easy to digest.
My instructor was unclear with the derivation of the formulas, so I've been having trouble with variation of parameters until I came across your video! Thank you so much!! :D
You are Great instructor . Thank you for appreciation the importance of the proof based learning and the the ideas behind the formulas and theories . I read that on Tenenbaum text and hope that my prof at class teach it , but for sorry he didn't . For that reason i believe in self learning and don't like the class approach about teaching us the final formulas or how to do that without understanding the underline information and details behind .
I can understand that condition 1 makes the calculations easier, however what I do not understand is: how this does not alter the solution if a different condition were chosen. Here is my guess: The particular solution to the differential equation is completly unique and therefore there can not be a loss of generality. If that is the case, we can choose a condition as we wish, as long as it does not contradict any of the other conditions set. Is this correct?
That is correct. He can choose any condition(s) he wants, and the V(1,2) will change accordingly to satisfy them. He will choose conditions that make the DE solvable in the best way possible, thus the conditions he chose, which are standard and generally rote learned.
In beginning of step 2, what inspires you to think the paricular solution will be v1 times y1 + v2 times y2? That is ... I wonder where did the premise of the nature of the particular solution come from? In other words, what makes you think there are such v1 and v2?
Thats the reason the method is called variation of parameters. Since yh = c1y1 + c2y2, lets assume parameters c1 and c2 are not constant although we can construct the function f(t) by "varying" or making them "variables". This is, change c1 to v1 and c2 to v2. Therefore, yp = v1y1+v2y2. The idea is deeply rooted in linear algebra, wherein you are using y1 and y2 as a vector basis in the space of real, well-behaved, functions like f(t). For this purpose, as indicated, f(t) must be an exponential, polynomial, sine or cosine function. This means, have R as domain and image. I think this is called a Hilbert space.
At 5:25 how can you set it equal to zero? I don't understand it. I understand why you want to set it equal to zero but don't how can it be possible to set random stuffs equal to 0?
Every solution to the differential equation are a combination of null solutions and particular solutions that come from those. We could impose any assumption that lets us get a particular solution because one particular solution is enough.(I wrote this to make my thoughts clearer.)
The Wronskian is just the matrix of coefficients for the linear equation the end. If you solve the two equation system with linear algebra methods, then you will be using the Wronskian. The Wronskian is not a shortcut for what he showed here, it's a result of it.
It's the idea of a superposition of differential equations. So a linear combination of a set of solutions will yield another solution that also satisfies the original differential equation.
@@geraldhuang7858 no, yp is not a linear combination of the set of homogenous solutions, otherwise yp would solve the homogenous equation too, which is not the case (yp solves the non-homogenous equation). Here we multiply y1 and y2 by two FUNCTIONS of x v1 and v2. The thing is we chose v1 and v2 so that it satisfies the equality yp = v1y1 + v2y2 and also the condition y1v1' + y2v2' = 0. And that is always possible because with these two functions v1 and v2 that we can chose, we have two degrees of freedom, so we can satisfy two conditions.
Yes. The Wronskian will be a 3x3 matrix, with original functions, derivatives, and second derivatives. Suppose our differential equation has the form: y"'(t) + b*y"(t) + c*y'(t) + d = g(t) Given yh1(t), yh2(t), and yh3(t) as the three fundamental solutions to the Homogeneous differential equation, the particular solution is: yp(t) = integral W1/W * g(t) dt + integral W2/W * g(t) dt + integral W3/W * g(t) dt W is the main Wronskian, with yh1(t), yh2(3) and yh3(t) as its first row, the derivatives as row 2, and the 2nd derivatives as row 3. W1, W2, and W3 are what I call, the Cramer Wronskians, because the theory behind them comes from Cramer's rule. To get W1, replace its first column with the column vector . To get W2, replace the 2nd column with the same column vector, and restore the first column. To get W3, replace the 3nd column with the same column vector, and restore the second column
just to verify, condition 1 would always apply using this method correct? since we didnt need to go further to derive the v's to their 2nd derivatives ? its nice to not derive everything to its second order derivatives with that condition to make it less painful. Thanks for a great explanation!
The function of the dependent variable (usually t) on the right, has to be one of the simple functions that either reduce to zero, or loop back as constant multiples of themselves when differentiated, in order to use undetermined coefficients. Polynomial functions reduce to zero, while exponentials, simple trig, and simple hyperbolic trig, will loop back as constant multiples of themselves. Anything else, that becomes significantly different after a calculus operation, like tangents, secants, and logs, will require variation of parameters.
The lighting, camera angle create a very poor, low-quality visual image. The over-exposed image makes it difficult to see what is written on the board. Other math channel videos are much more professional than this garbage
I actually understood this! you got me through cal 1, cal 2, and now differential equations. Thank you!
4:30 you should explain why you can impose that condition.
And so we reach the part of this topic that seemingly no one explains
@@JDMaxton1999 Basically, it's not so much of an assumption, but more like an "educated guess". Whomever figured this out realized that making y1v1' + y2v2' = 0 would cut our problem in half, and has a nice form to plug into the DE. You can use any random number as a guess, but what you end up with is some nasty or potentially unsolvable DE problems. So this is a nice way to find a general form. Like BPRP was saying in his video, if we kept the first derivative as is, then when we plug it back into our original DE, we will be running into a second order DE within a second order DE. Just nasty, nasty stuff.
Remember, c1y1 + c2y2 is A fundamental set, not THE fundamental set. There is potentially more than one answer for these problems based upon what initial conditions are given to either the constants or what f(t) is. This is just THE fundamental set based upon the given conditions.
I hoped this helps. I'm also not a fan of "hand wave, black box magic" with math, but sometimes it's better to trust the system then learn later on why. I find this to be the "if your quadratic is becoming unsolvable, see if you can graph it and get an answer that way" equivalent in DE.
The explanation is given by the *second* answer to the math stackexchange question titled "Justifying an Assumption Made While Deriving the Method of Variation of Parameters". It explains the *simple* reason why 'osculating functions' will always satisfy this constraint. I have no idea why this step is never justified, now that I know the answer is so easy to digest.
You're one of the reasons why I currently have an A in my differential equations class. Thank you!!!
"one of the reasons" you mean you watch someone else videos also??? You traitor😠😂
hi Brain
@@MrEngineer-TRD Hahaha, oddly enough Ik a guy with your name. Are you a civil engineering uofa student?
@@abdallababikir4473 University of Alberta in Canada?
@@boyhandsome4047 yeah. I asked my friend who has this name but he's not the guy to make this comment.
My instructor was unclear with the derivation of the formulas, so I've been having trouble with variation of parameters until I came across your video!
Thank you so much!! :D
Ive watched many videos, and this is one if the few that went into explaining and proving the methods and formulas for solving this problem
You are Great instructor . Thank you for appreciation the importance of the proof based learning and the the ideas behind the formulas and theories . I read that on Tenenbaum text and hope that my prof at class teach it , but for sorry he didn't . For that reason i believe in self learning and don't like the class approach about teaching us the final formulas or how to do that without understanding the underline information and details behind .
I can understand that condition 1 makes the calculations easier, however what I do not understand is: how this does not alter the solution if a different condition were chosen.
Here is my guess: The particular solution to the differential equation is completly unique and therefore there can not be a loss of generality. If that is the case, we can choose a condition as we wish, as long as it does not contradict any of the other conditions set.
Is this correct?
Now I got the reason why condition 1 makes sense. thx!
That is correct. He can choose any condition(s) he wants, and the V(1,2) will change accordingly to satisfy them. He will choose conditions that make the DE solvable in the best way possible, thus the conditions he chose, which are standard and generally rote learned.
You make math more interesting, THANK YOU!
this guy is so happy it makes me happy
In beginning of step 2, what inspires you to think the paricular solution will be v1 times y1 + v2 times y2?
That is ... I wonder where did the premise of the nature of the particular solution come from?
In other words, what makes you think there are such v1 and v2?
The formula for yp is proven in the book
Thats the reason the method is called variation of parameters. Since yh = c1y1 + c2y2, lets assume parameters c1 and c2 are not constant although we can construct the function f(t) by "varying" or making them "variables". This is, change c1 to v1 and c2 to v2. Therefore, yp = v1y1+v2y2. The idea is deeply rooted in linear algebra, wherein you are using y1 and y2 as a vector basis in the space of real, well-behaved, functions like f(t). For this purpose, as indicated, f(t) must be an exponential, polynomial, sine or cosine function. This means, have R as domain and image. I think this is called a Hilbert space.
Sir. Why did we assume that (y1v-1 + y2v-2) equals zero ....? !!!
It is necessary to assume something otherwise you cant solve the differential equation
You just searxh for a particular soloution so any soloution ,and thus you look for one that satsfies that condition to simplify stufd
Oh my god. You are my professor! You have crazy math technique
At 5:25 how can you set it equal to zero? I don't understand it. I understand why you want to set it equal to zero but don't how can it be possible to set random stuffs equal to 0?
it's an assumption
All the problems that we solve should also obey that assumption. Then only this method can be used there.
Every solution to the differential equation are a combination of null solutions and particular solutions that come from those. We could impose any assumption that lets us get a particular solution because one particular solution is enough.(I wrote this to make my thoughts clearer.)
Finally some reasoning to the equations! Thanks for clarifying the process.
You should also introduced the shortcut called Wronskian Method. It's very effective for Variation of Parameters.
The Wronskian is just the matrix of coefficients for the linear equation the end. If you solve the two equation system with linear algebra methods, then you will be using the Wronskian. The Wronskian is not a shortcut for what he showed here, it's a result of it.
yea
Yep, Wronskian is not for here
How did we come up with the condition that y_1 v_1' + vy_2 v_2' =0 ?
Best explanation I've seen so far
I love you
This guy is a legend
Thank you @blackpenredpen, you solved my problem !
A very nice explanation. Thank you very much.
My pleasure
great video ! Why are we allowed to impose condition 1 ?
Holy moly! All I have to do is use formulas!!
Very nice video! Thank you so much!
So useful! Thank you!
Nice video!
To use this method for higher order differential equation (n>2) we just impose more conditions?
Amazing video thank you so much
Perfect explanation
The video sound is pretty good, beyond my imagination
I got a question actually.Will Condition 1 always be satisfied??
It's an arbitrarily imposed constraint so that you do not have to deal with second order y terms.
Very good explanation.
Thanks
Also -- Thank you for the excellent demonstration!
Awesome, you have been a lot of help.
Awesome sir.
Talk of people who are genius talk of this guy
Sorry I'm not as fast of a learner, but why did you set yp = v1y1 + v2y2?
It's the idea of a superposition of differential equations. So a linear combination of a set of solutions will yield another solution that also satisfies the original differential equation.
@@geraldhuang7858 no, yp is not a linear combination of the set of homogenous solutions, otherwise yp would solve the homogenous equation too, which is not the case (yp solves the non-homogenous equation). Here we multiply y1 and y2 by two FUNCTIONS of x v1 and v2. The thing is we chose v1 and v2 so that it satisfies the equality yp = v1y1 + v2y2 and also the condition y1v1' + y2v2' = 0. And that is always possible because with these two functions v1 and v2 that we can chose, we have two degrees of freedom, so we can satisfy two conditions.
Thank u so much 🔥🔥🔥🔥😎😎😎
Does this idea also work for third order differential non-homogenous differential equations?
Yes. The Wronskian will be a 3x3 matrix, with original functions, derivatives, and second derivatives.
Suppose our differential equation has the form:
y"'(t) + b*y"(t) + c*y'(t) + d = g(t)
Given yh1(t), yh2(t), and yh3(t) as the three fundamental solutions to the Homogeneous differential equation, the particular solution is:
yp(t) = integral W1/W * g(t) dt + integral W2/W * g(t) dt + integral W3/W * g(t) dt
W is the main Wronskian, with yh1(t), yh2(3) and yh3(t) as its first row, the derivatives as row 2, and the 2nd derivatives as row 3.
W1, W2, and W3 are what I call, the Cramer Wronskians, because the theory behind them comes from Cramer's rule.
To get W1, replace its first column with the column vector .
To get W2, replace the 2nd column with the same column vector, and restore the first column.
To get W3, replace the 3nd column with the same column vector, and restore the second column
Thank you Sir
just to verify, condition 1 would always apply using this method correct? since we didnt need to go further to derive the v's to their 2nd derivatives ? its nice to not derive everything to its second order derivatives with that condition to make it less painful. Thanks for a great explanation!
to understand this guy come prepare, read prior
Thank you so much
is there a solution to ay"+by'y=0?
Help me with these problems please dy/dx=y^6_2x^2/2xy^5+x^2y^2 using homogenous method
Feel like I went back 10 years since I got used to the beard
please explain .....In
which conditions cannot we use undetermined coefficients methods ...
The function of the dependent variable (usually t) on the right, has to be one of the simple functions that either reduce to zero, or loop back as constant multiples of themselves when differentiated, in order to use undetermined coefficients. Polynomial functions reduce to zero, while exponentials, simple trig, and simple hyperbolic trig, will loop back as constant multiples of themselves.
Anything else, that becomes significantly different after a calculus operation, like tangents, secants, and logs, will require variation of parameters.
I just forgot what i asked. :) Anyway Thanks indeed@@carultch
Why such a condition 1 ?
I could not apply that for u'+u=x, why?
That would be a first order linear DE. You can use the integrating factor approach (wich is simpler). :)
that’s treasure
nice
thank you
Ah yes! Now I see!
Anda boleh memilih untuk salah satu daripada hadiah di atas
for third order?
Sick! i love math
legend
if this doesn't make you love maths i dunno what will :D
thank u verye much
7:01 we do mind 😂
This was godtier.
god bless u
❤️❤️❤️❤️
haha that music at the end is ridiculous. great video nonetheless
The lighting, camera angle create a very poor, low-quality visual image. The over-exposed image makes it difficult to see what is written on the board. Other math channel videos are much more professional than this garbage
why are you here, give me their channel links
His English is terrible-- poor pronunciation, enunciation, diction, & elocution. He needs to work on articulation.