Sir may you and your family live happily forever. such quality material you are providing us for free....Damnn this is so much better than the paid courses...Thanks a lot sir.
As an Electrical Engineering sophomore struggling through differential equations, thank you for making these videos! They were also very helpful for my Calc 3 class last year. This is by far my favorite math channel on UA-cam, your explanations are great!
well , it is all ancient mathematics . If Euler were still alive he could show us how he did it . The best book is : Swokowski's Caclulus 5th Edition , see page 1028 1029 Spoiler alert , the book shows why P(x) seems to disappear from the final form, after multiplying with the Integration Factor : e^-x^3 dy/dx - ( 3x^2 )ye^-x^3 = ( x^2 )e^-x^3 and equivalently : Dx ( ye^-x^3 ) = ( x^2 )e^-x^3 ( where did factor 3x^2 go ? , aka P(x) )
u maths teacher from other countries are literally awesome and stand much above our indian teachers in explaining. Thanx for such a beautiful explaination
Oh believe Clash, they're not all like this over here. I'm at an American University, but I'm coming to UA-cam University to learn about it just like you are my friend. We're lucky to have this.
@@captaingerbil1234 The professor is Canadian. Incidentally, he lives in a city in Canada that enjoys noticeably milder winters than you do in Washington, DC! So, it would be nice if you went to his university in Canada!
Sir I just want to thank you so much for doing these videos. I am studying engineering and because of covid we started a lot later this year and as a result rushed through a lot of the beginning stuff. Because of this I was struggling to keep up and struggling to build on my very 'un-established' foundations of differentials. Your vids allow me to understand concepts better and rewind when I don't and I really really appreciate it. Love and appreciation from South Africa!!
This is absolutely fantabulous. Currently revising ODEs for PDEs and this series has done wonders. Especially this video on Integrating factors, very sound explanations all through. You've done a wonderful job and I hope you're aware that is highly appreciated.
Always great when you can find a video like this if your calculus textbook doesn't explain it clearly. Abstract concepts like this are hard for me to grasp but watching this only 3 times was enough for me to understand it perfectly. 10/10
It seems like you used "wishful thinking " when you said "I would love ..." It's one of my favorite problem solving strategies. I wish my teachers would have made this technique (and others) more explicit when I was in highschool/college. It makes motivating proofs easier for everyone. Loved the lecture. Got hooked by your calculus series.
I would like to add that there is another formulation of this idea which, while I admit it is potentially less intuitive, it is more easily generalizable to higher-order equations, and it shows the true connection that this subject has with linear algebra, which in turn, may make it more illuminating. Linear first-order equations can be written as D[y(x)] + p(x)·y(x) = q(x), as explained in the video, where D[y(x)] stands for the derivative of y. It will make sense why I am using this notation instead of the usual y' in just a moment, but bear with me. Notationally, you may be tempted to "factor out" the y from the right, writing this as [D + p(x)][y(x)] = q(x), and if you rename the object A := D + p(x), then you get an equation that looks like A[y(x)] = q(x). Now this looks a lot more like an equation you would encounter in linear algebra: y and q are functions, and A is some type of object that behaves like a linear operator acting on the space of differentiable functions, so in a meaningful sense, A is very much analogous to a matrix here. This representation makes it obvious what is it that you need to do to solve the equation: you want to "invert" the linear operator A, find some A^(-1), so that y(x) = [A^(-1)][q(x)] is the solution to the equation. For p(x) = 0 for almost all x, this is implies A := D, and so finding A^(-1) is trivial: you simply integrate using the initial conditions. However, for any other p(x), this is completely non-obvious. This is where the integration factors comes into play. It is not obvious how to invert an operator that looks like D + p(x), but if you could somehow reexpress A[y(x)] as [r(x)^(-1)·D][r(x)·y(x)], then this would make the problem trivial again. What this video does is precisely teach you that you can always do this: if r(x) = exp(Antiderivative[p(x)]), then you can always write D[r(x)·y(x)] = r(x)·q(x), which is indeed equivalent to r(x)^(-1)·D[r(x)·y(x)] = q(x). Why do I want to rewrite A[y(x)] as r(x)^(-1)·D[r(x)·y(x)]? Before I explain this, let me make one final change to the notation. Let the linear operator R be defined by the rule R[y(x)] = r(x)·y(x). Hence r(x)^(-1)·D[r(x)·y(x)] = [R^(-1)·D·R][y(x)] = q(x). Now it should become clear why I wanted to rewrite A[y(x)] as [R^(-1)·D·R][y(x)]: because this is just the same as saying that A = R^(-1)·D·R, where A, D, R are linear operators. Notice how this is exactly analogous to the diagonalization of a matrix into a diagonal eigenvalue matrix and an eigenvectit matrix. In effect, using integration factors in the study of differential is just a "diagonalization" of the operator A := D + p(x), which is the operator we want to invert. With this, solving the equation is now trivial, and the solution becomes y(x) = [R^(-1)·D^(-1)·R][q(x)] = 1/r(x)·Integral[r(x)·q(x)], which is exactly what we obtained in the video! Of course, I am being somewhat handwavy here, since technically, D is not an invertible operator in the ordinary sense, and so D^(-1) here represents integration with usage of a specific initial condition, but the core idea is still the same: solving a linear differential equation is just diagonalizing the operator A. It is, indeed, just linear algebra, and this is the hidden truth that I have been trying to uncover here in my explanation. This formulation is not only illuminating as to the linear-algebraic nature of these equations, but it is easy also useful, because it gives us a method by which you can solve linear equation of higher-order in terms of this first-order idea, as long as you are able to "factorize" the equation. What do I mean by this? As an example, suppose you have an equation y''(x) + f(x)·y'(x) + g(x)·y(x) = h(x), which, for reasons that now should have become apparent, I should rewrite as [D^2 + f(x)·D + g(x)][y(x)] = h(x). Again, with A := D^2 + f(x)·D + g(x), this is just A[y(x)] = h(x), a linear equation, and you want to "invert" A so that the solutions look like y(x) = [A^(-1)][h(x)]. Here, A is again a linear operator, but this time, it is equal to a quadratic polynomial in D instead of a first-degree polynomial in D. Here is the inspiration: if polynomials with complex coefficients can be always factored into a product of polynomials of first-degree with complex coefficients again, then, should we not be able to do the same thing with polynomials in D with functional coefficients? The answer is yes, with a large caveat: this "multiplication" of linear operators, which are comprised of sums of products of D and functions, is not commutative. This is to say, R·D is not the same as D·R: again, entirely analogous to matrix multiplication in linear algebra. Thus, the order in which you do the factorization matters, and this can also complicate things. To see this explicitly, you can carefully evaluate {[D - r(x)]·[D - s(x)]}[y(x)] as [D - r(x)]{D[y(x)] - s(x)·y(x)} = D{D[y(x)] - s(x)·y(x)} - r(x)·D[y(x)] + r(x)·s(x)·y(x) = (D^2)[y(x)] - D[s(x)·y(x)] - r(x)·D[y(x)] + r(x)·s(x)·y(x) = (D^2)[y(x)] - D[s(x)]·y(x) - s(x)·D[y(x)] - r(x)·D[y(x)] + r(x)·s(x)·y(x) = {D^2 - [r(x) + s(x)]·D + [r(x)·s(x) - s'(x)]}[y(x)]. This gives you the factorization D^2 - [r(x) + s(x)]·D + [r(x)·s(x) - s'(x)] = [D - r(x)]·[D - s(x)], and the lack of commutativity is manifested in the asymmetric expression r(x)·s(x) - s'(x). Anyhow, the idea is that, in factorizing the quadratic polynomial A as [D - r(x)]·[D - s(x)], you can write A as R1^(-1)·D·R1·R2^(-1)·D·R2, where R2 is the operator that multiplies its input by the function exp(-Antiderivative[s(x)]), and R1 is the operator that multiplies its input by the function exp(-Antiderivative[r(x)]). The same idea applies for higher-order equations, where you factorize higher-degree polynomials in D. What does this mean? It means that solving any linear equation, in theory, merely reduces to multiplying by an appropriate integration factor, integrating, and repeating the process, and this can be done in the other direction too, by simply substituting y(x) with the appropriate t(x)·y(x) for some factor t, and proceeding from there. You may not even need to factor the equation, you only need to know that you can always find the appropriate integration factor, because this idea guarantees its existence. This idea is what makes linear equations so much simpler to solve than non-linear ones in general. This idea also opens the door to the discipline of mathematical study known as operator theory, where a treatment of rigor is given to these ideas of linear operators, expanding these concepts beyond the linear algebra of matrices and Euclidean R^n spaces. This turns out to have significant applications in the sciences, especially in quantum physics, but it also useful fir the study of other disciplines in mathematics, in turn.
As an addendum, you may be wondering how would you go about factorizing D^2 + f(x)·D + g(x), where f and g are given, into [D - r(x)]·[D - s(x)], where r and s are the unknown functions. Since [D - r(x)]·[D - s(x)] = D^2 - [r(x) + s(x)]·D + [r(x)·s(x) - s'(x)], this gives you the system of equations f(x) = -[r(x) + s(x)] and r(x)·s(x) - s'(x) = g(x). The first equation implies r(x) = -[f(x) + s(x)], which implies the second equation is just equivalent to -s(x)·[s(x) + f(x)] - s'(x) = g(x), which is equivalent to s'(x) = -s(x)·[s(x) + f(x)] - g(x)] = -[s(x)^2 + f(x)·s(x) + g(x)]. This is, in fact, the Riccati differential equation, which deserves separate treatment in its own right, but is solvable.
Hi Dr. Trefor, I'm told from the grapevine that the (first-order linear) ODEs which are amenable to the method of integrating factors are in fact 'non-exact' diff eqs that can be turned into exact diff eqs precisely from this integrating factor. If there's no misconception there or if I'm missing anything out, then this picture also leans heavily on the math of differential 1-forms, which have an isomorphism with vector fields, and it turns out exact diff. forms have a correspondence with conservative vector fields. (and I'd assume vice versa: non-exact diff forms correspond to non-conservative vector fields) In that case then with that geometric picture/correspondence in mind, since we have that exact diff eqs/exact diff forms conservative vector field picture in mind, then by turning a non-exact diff eq./form into an exact one via these integrating factors, aren't we dually in the process also turning a non-conservative vector field into a conservative one? I'd like to know more about this potential (no pun intended) geometric correspondence.
This was posted a year ago 😂and I can't help but thank God cause it's helped me so much 😂all our teacher said was to memorize and it really didn't make sense but you 😂😂you came like a hero 🔥💯
Note that the integrals along the way generate a few constants, but they all end up absorbed in the final constant anyway, that's why they weren't mentioned. The sign when we get rid of the absolute value in e ^ ln |r(x)| is ultimately also absorbed into the constants of integration.
Hi Dr. Bazett! Thank you for the videos. Not sure how to put this, it is really not personal, but is there a version without you appearing in the videos? The gesticulation is sometimes overwhelming and it takes away focus from the material presented. Thank you.
Thank you! I actually do hope to do a real analysis series. Perhaps not a completely normal one, but one that really showcases how bizarre and strange the real numbers are!
Question about a step at time 7:36 Can I get some justification on why r'(x) = 1? Assuming we only consider r'(x) = mx + b. Shouldn't we consider the m? Also with hindsight, we know that r(x) contains an exponential. This seems to conflict with the assumption that we could solve Int( r'x)/r(x) ) by simplifying it to Int( 1/r(x) ) to then find Int( p(x) ) = ln| r(x) | . we seem to be treating r as a variable and not as a function between step 1 and step 3. I really like the video I just get hung up on things like this and get distracted from doing my work.
I feel very stupid for asking this... around 8:00 mark, wouldnt the left hand side integrat to ln(r(x)/r(x)) = 1? Because the integral of r'(x) is r(x)... wait but in that case you would have to use quotient rule. Now i am very confused by this notation.
Hey, do the Differential Equations playlists cover the same content as the content in the book: W.E. Boyce, R.C.DiPrima, Elementary differential equations and boundary value problems, Wiley, New York?
Hi. This is a nice video. However, while you explained that there is no constant of integration needed in the first integral you did (to find the integrating factor), the second integral you do does need to have a constant of integration. Then when you divide by the integrating factor at the end, that term becomes a non-trivial function of x. Or am I mistaken?
Why the ode would look like that at 1:51 if it's a linear ode? ... Is there a reason behind the ode to look like that or is that a notation we take or assume?
It is just notation, the important part is that the coefficients of y or its derivatives are just any old function in terms of ONLY the independent variable. So x^2y'+cos(x)y=3 is linear because the coefficients of y and y' don't depend on y and y'.
Hey i really appreciate your videos, way better explained than the average university prof. But your textbook link does not work. Thought id mention. Your website since Uvic also is not accessible for general public.
If you set C=0 for this setup, you'll just end up with a constant of 1 in front of the exponential. Since this constant will show up both above and below a division bar in the final answer, it doesn't matter what you make your +C. Because e^C will show up both above and below the division bar, and will cancel out of the equation. There is an integral where it will matter, and where we'll ultimately solve for +C with an initial condition. But usually when there are more integrals in a solution, than there are orders of differentiation in the original equation, only as many arbitrary constants as the order of differentiation will need remain by end of your general solution, to solve for with initial conditions.
Sir may you and your family live happily forever. such quality material you are providing us for free....Damnn this is so much better than the paid courses...Thanks a lot sir.
So nice of you!
As an Electrical Engineering sophomore struggling through differential equations, thank you for making these videos! They were also very helpful for my Calc 3 class last year. This is by far my favorite math channel on UA-cam, your explanations are great!
Great to hear!
Sometimes I wonder why we can't get this type of professor in our college and school?
haha, thank you!
well , it is all ancient mathematics .
If Euler were still alive he could show us how he did it .
The best book is : Swokowski's Caclulus 5th Edition , see page 1028 1029
Spoiler alert , the book shows why P(x) seems to disappear
from the final form, after multiplying with the Integration Factor :
e^-x^3 dy/dx - ( 3x^2 )ye^-x^3 = ( x^2 )e^-x^3
and equivalently :
Dx ( ye^-x^3 ) = ( x^2 )e^-x^3 ( where did factor 3x^2 go ? , aka P(x) )
u maths teacher from other countries are literally awesome and stand much above our indian teachers in explaining.
Thanx for such a beautiful explaination
Oh believe Clash, they're not all like this over here. I'm at an American University, but I'm coming to UA-cam University to learn about it just like you are my friend. We're lucky to have this.
@@captaingerbil1234 The professor is Canadian. Incidentally, he lives in a city in Canada that enjoys noticeably milder winters than you do in Washington, DC! So, it would be nice if you went to his university in Canada!
Sir I just want to thank you so much for doing these videos. I am studying engineering and because of covid we started a lot later this year and as a result rushed through a lot of the beginning stuff. Because of this I was struggling to keep up and struggling to build on my very 'un-established' foundations of differentials. Your vids allow me to understand concepts better and rewind when I don't and I really really appreciate it. Love and appreciation from South Africa!!
Thanks so much!!
Please make a video on tensors used in general relativity 🙏🙏💓
Sir you and Grant Sanderson(3b1b) Sir are the persons who made me love the subject which I hated the most.❤️thank you so much....Love from India
Sir , I have a question, do you use the licensed version of Geogebra ,since your videos contain ads.
Thanks a lot for the heart, sir❤️❤️
This is absolutely fantabulous. Currently revising ODEs for PDEs and this series has done wonders. Especially this video on Integrating factors, very sound explanations all through. You've done a wonderful job and I hope you're aware that is highly appreciated.
Always great when you can find a video like this if your calculus textbook doesn't explain it clearly. Abstract concepts like this are hard for me to grasp but watching this only 3 times was enough for me to understand it perfectly. 10/10
It seems like you used "wishful thinking " when you said "I would love ..." It's one of my favorite problem solving strategies. I wish my teachers would have made this technique (and others) more explicit when I was in highschool/college. It makes motivating proofs easier for everyone. Loved the lecture. Got hooked by your calculus series.
haha proof by wishful thinking is my favourite proof method, particularly because it so often gives hints as to the right method!
This was literally going to be my comment on this video...
I would like to add that there is another formulation of this idea which, while I admit it is potentially less intuitive, it is more easily generalizable to higher-order equations, and it shows the true connection that this subject has with linear algebra, which in turn, may make it more illuminating.
Linear first-order equations can be written as D[y(x)] + p(x)·y(x) = q(x), as explained in the video, where D[y(x)] stands for the derivative of y. It will make sense why I am using this notation instead of the usual y' in just a moment, but bear with me. Notationally, you may be tempted to "factor out" the y from the right, writing this as [D + p(x)][y(x)] = q(x), and if you rename the object A := D + p(x), then you get an equation that looks like A[y(x)] = q(x). Now this looks a lot more like an equation you would encounter in linear algebra: y and q are functions, and A is some type of object that behaves like a linear operator acting on the space of differentiable functions, so in a meaningful sense, A is very much analogous to a matrix here. This representation makes it obvious what is it that you need to do to solve the equation: you want to "invert" the linear operator A, find some A^(-1), so that y(x) = [A^(-1)][q(x)] is the solution to the equation. For p(x) = 0 for almost all x, this is implies A := D, and so finding A^(-1) is trivial: you simply integrate using the initial conditions. However, for any other p(x), this is completely non-obvious.
This is where the integration factors comes into play. It is not obvious how to invert an operator that looks like D + p(x), but if you could somehow reexpress A[y(x)] as [r(x)^(-1)·D][r(x)·y(x)], then this would make the problem trivial again. What this video does is precisely teach you that you can always do this: if r(x) = exp(Antiderivative[p(x)]), then you can always write D[r(x)·y(x)] = r(x)·q(x), which is indeed equivalent to r(x)^(-1)·D[r(x)·y(x)] = q(x). Why do I want to rewrite A[y(x)] as r(x)^(-1)·D[r(x)·y(x)]? Before I explain this, let me make one final change to the notation. Let the linear operator R be defined by the rule R[y(x)] = r(x)·y(x). Hence r(x)^(-1)·D[r(x)·y(x)] = [R^(-1)·D·R][y(x)] = q(x). Now it should become clear why I wanted to rewrite A[y(x)] as [R^(-1)·D·R][y(x)]: because this is just the same as saying that A = R^(-1)·D·R, where A, D, R are linear operators. Notice how this is exactly analogous to the diagonalization of a matrix into a diagonal eigenvalue matrix and an eigenvectit matrix. In effect, using integration factors in the study of differential is just a "diagonalization" of the operator A := D + p(x), which is the operator we want to invert. With this, solving the equation is now trivial, and the solution becomes y(x) = [R^(-1)·D^(-1)·R][q(x)] = 1/r(x)·Integral[r(x)·q(x)], which is exactly what we obtained in the video! Of course, I am being somewhat handwavy here, since technically, D is not an invertible operator in the ordinary sense, and so D^(-1) here represents integration with usage of a specific initial condition, but the core idea is still the same: solving a linear differential equation is just diagonalizing the operator A. It is, indeed, just linear algebra, and this is the hidden truth that I have been trying to uncover here in my explanation.
This formulation is not only illuminating as to the linear-algebraic nature of these equations, but it is easy also useful, because it gives us a method by which you can solve linear equation of higher-order in terms of this first-order idea, as long as you are able to "factorize" the equation. What do I mean by this? As an example, suppose you have an equation y''(x) + f(x)·y'(x) + g(x)·y(x) = h(x), which, for reasons that now should have become apparent, I should rewrite as [D^2 + f(x)·D + g(x)][y(x)] = h(x). Again, with A := D^2 + f(x)·D + g(x), this is just A[y(x)] = h(x), a linear equation, and you want to "invert" A so that the solutions look like y(x) = [A^(-1)][h(x)]. Here, A is again a linear operator, but this time, it is equal to a quadratic polynomial in D instead of a first-degree polynomial in D. Here is the inspiration: if polynomials with complex coefficients can be always factored into a product of polynomials of first-degree with complex coefficients again, then, should we not be able to do the same thing with polynomials in D with functional coefficients? The answer is yes, with a large caveat: this "multiplication" of linear operators, which are comprised of sums of products of D and functions, is not commutative. This is to say, R·D is not the same as D·R: again, entirely analogous to matrix multiplication in linear algebra. Thus, the order in which you do the factorization matters, and this can also complicate things. To see this explicitly, you can carefully evaluate {[D - r(x)]·[D - s(x)]}[y(x)] as [D - r(x)]{D[y(x)] - s(x)·y(x)} = D{D[y(x)] - s(x)·y(x)} - r(x)·D[y(x)] + r(x)·s(x)·y(x) = (D^2)[y(x)] - D[s(x)·y(x)] - r(x)·D[y(x)] + r(x)·s(x)·y(x) = (D^2)[y(x)] - D[s(x)]·y(x) - s(x)·D[y(x)] - r(x)·D[y(x)] + r(x)·s(x)·y(x) = {D^2 - [r(x) + s(x)]·D + [r(x)·s(x) - s'(x)]}[y(x)]. This gives you the factorization D^2 - [r(x) + s(x)]·D + [r(x)·s(x) - s'(x)] = [D - r(x)]·[D - s(x)], and the lack of commutativity is manifested in the asymmetric expression r(x)·s(x) - s'(x). Anyhow, the idea is that, in factorizing the quadratic polynomial A as [D - r(x)]·[D - s(x)], you can write A as R1^(-1)·D·R1·R2^(-1)·D·R2, where R2 is the operator that multiplies its input by the function exp(-Antiderivative[s(x)]), and R1 is the operator that multiplies its input by the function exp(-Antiderivative[r(x)]). The same idea applies for higher-order equations, where you factorize higher-degree polynomials in D. What does this mean? It means that solving any linear equation, in theory, merely reduces to multiplying by an appropriate integration factor, integrating, and repeating the process, and this can be done in the other direction too, by simply substituting y(x) with the appropriate t(x)·y(x) for some factor t, and proceeding from there. You may not even need to factor the equation, you only need to know that you can always find the appropriate integration factor, because this idea guarantees its existence.
This idea is what makes linear equations so much simpler to solve than non-linear ones in general. This idea also opens the door to the discipline of mathematical study known as operator theory, where a treatment of rigor is given to these ideas of linear operators, expanding these concepts beyond the linear algebra of matrices and Euclidean R^n spaces. This turns out to have significant applications in the sciences, especially in quantum physics, but it also useful fir the study of other disciplines in mathematics, in turn.
As an addendum, you may be wondering how would you go about factorizing D^2 + f(x)·D + g(x), where f and g are given, into [D - r(x)]·[D - s(x)], where r and s are the unknown functions. Since [D - r(x)]·[D - s(x)] = D^2 - [r(x) + s(x)]·D + [r(x)·s(x) - s'(x)], this gives you the system of equations f(x) = -[r(x) + s(x)] and r(x)·s(x) - s'(x) = g(x). The first equation implies r(x) = -[f(x) + s(x)], which implies the second equation is just equivalent to -s(x)·[s(x) + f(x)] - s'(x) = g(x), which is equivalent to s'(x) = -s(x)·[s(x) + f(x)] - g(x)] = -[s(x)^2 + f(x)·s(x) + g(x)]. This is, in fact, the Riccati differential equation, which deserves separate treatment in its own right, but is solvable.
Thank you for sharing!
I tried to read but I couldnt HAHAHA. Thanks for sharing tho, I love to see more generalized solutions to problems!
Wow...thank you so much...I think nobody else could be this much clearer
Hi Dr. Trefor, I'm told from the grapevine that the (first-order linear) ODEs which are amenable to the method of integrating factors are in fact 'non-exact' diff eqs that can be turned into exact diff eqs precisely from this integrating factor.
If there's no misconception there or if I'm missing anything out, then this picture also leans heavily on the math of differential 1-forms, which have an isomorphism with vector fields, and it turns out exact diff. forms have a correspondence with conservative vector fields. (and I'd assume vice versa: non-exact diff forms correspond to non-conservative vector fields)
In that case then with that geometric picture/correspondence in mind, since we have that exact diff eqs/exact diff forms conservative vector field picture in mind, then by turning a non-exact diff eq./form into an exact one via these integrating factors, aren't we dually in the process also turning a non-conservative vector field into a conservative one?
I'd like to know more about this potential (no pun intended) geometric correspondence.
How cool. The key part for the development of the method is the dot product of the derivative ! Thank you
This was posted a year ago 😂and I can't help but thank God cause it's helped me so much 😂all our teacher said was to memorize and it really didn't make sense but you 😂😂you came like a hero 🔥💯
Note that the integrals along the way generate a few constants, but they all end up absorbed in the final constant anyway, that's why they weren't mentioned.
The sign when we get rid of the absolute value in e ^ ln |r(x)| is ultimately also absorbed into the constants of integration.
You are a stellar teacher. Thank you for all the help you have given me.
i've had integrating factors explained to me 3 times now and it always impresses me haha
great video thank you
Professor Bazett, thank you for an excellent analysis and derivation of Linear Differential Equations and the classical Method of Integrating Factors.
You always know what we confuse about. You are a great and amazing teacher.
Such a great video, thank you for explaining the methodology in a clear way!
Glad it was helpful!
Kind of complicated but i think solving some examples will get me grasp it more
Thank you very much, these videos are really helpful for my exam preparation, greetings from Eastern Europe
Great great explanation!! Thank you!!!
Great video. Thank you
Wow thank you so much this is the best explanation I've ever seen
By far the clearest explanation
You deserve a medal
thank you for the videos! really enjoy watching your explanations, in addition to my text books.
Very grateful for this video
Your explanation is VERY clear, THANKS A LOT!
8:13 due to ln formula d/DX y = Y'/y 10:00 substitution , e both base
please upload the full playlist as soon as possible.
Should be coming out 2-3 per week now:)
Hi Dr. Bazett! Thank you for the videos. Not sure how to put this, it is really not personal, but is there a version without you appearing in the videos? The gesticulation is sometimes overwhelming and it takes away focus from the material presented. Thank you.
perfect sir thank you.go on
Thank you!!
Perfect explanation!
Hello Sir.. I want to mention this, Your videos are awesome and helpful to many🙏
Lots of love from India❤️
Tx for the lecture 🙏
will you also make lecture on real analysis (request)
The way you teach is unique
Thank you! I actually do hope to do a real analysis series. Perhaps not a completely normal one, but one that really showcases how bizarre and strange the real numbers are!
Wow, so clear. Watching from Grenoble, France.
Please, notify me when you write your maths textbook
Thank you! Check out the link in the description for the ODE text:)
Sending lots of respect and love.🤗
Your channel is wonderful and useful. All the support. I am from Egypt. Excuse me for the inaccurate language 😂❤
Great explanation!
Question about a step at time 7:36
Can I get some justification on why r'(x) = 1?
Assuming we only consider r'(x) = mx + b.
Shouldn't we consider the m?
Also with hindsight, we know that r(x) contains an exponential.
This seems to conflict with the assumption that we could solve Int( r'x)/r(x) )
by simplifying it to Int( 1/r(x) ) to then find Int( p(x) ) = ln| r(x) | .
we seem to be treating r as a variable and not as a function between step 1 and step 3.
I really like the video I just get hung up on things like this and get distracted
from doing my work.
You are mistaken.😅
[ln|r(x)|]' = r'(x)/r(x) 👈the chain rule
So that's why
[ln|x|]' = x'/x = 1/x
😉
Nice explation of i.factor. Thanks
Hi sir plz explain some example sums on existence and uniqueness solution for I.V.P on D.E
I feel very stupid for asking this... around 8:00 mark, wouldnt the left hand side integrat to ln(r(x)/r(x)) = 1? Because the integral of r'(x) is r(x)... wait but in that case you would have to use quotient rule. Now i am very confused by this notation.
Thank u so much for teach us
Thank you sir!!
King 👑
Oh my God this guy is good
I loved it man. Thanks for helping
You are great, tyvm!
Can you explain why the integral of r'(x)/r(x) with respect to x is r(x)? I don't see how
how are you dividing by rx to integrate? That only happens with exponentials, and rx * y is not an exponential.
Hey, do the Differential Equations playlists cover the same content as the content in the book: W.E. Boyce, R.C.DiPrima, Elementary differential equations and boundary value problems, Wiley, New York?
I mainly used Boyce diprima but both texts are similar
excellent.
Hi. This is a nice video. However, while you explained that there is no constant of integration needed in the first integral you did (to find the integrating factor), the second integral you do does need to have a constant of integration. Then when you divide by the integrating factor at the end, that term becomes a non-trivial function of x. Or am I mistaken?
That’s exactly right.
Really amazing
Very good!
Why the ode would look like that at 1:51 if it's a linear ode? ... Is there a reason behind the ode to look like that or is that a notation we take or assume?
It is just notation, the important part is that the coefficients of y or its derivatives are just any old function in terms of ONLY the independent variable. So x^2y'+cos(x)y=3 is linear because the coefficients of y and y' don't depend on y and y'.
Plz make a video about the bilinear form.
Thank you very much
You're most welcome!
why would it not be r'(x) ln(r(x)) = integral of p (x)? what happens to the r'(x) when we integrate?
Why is the integral of r'(x)/r(x) equal to ln(r(x))? Is this always true?
Try taking the derivative of ln(r(x)), and recall integration undoes differentiation.
Excuse me, what is the solution of xy'+y= 2x, it is my assignment, please solve me
Excellent 👏
after all these videos i'll owe you a part of my salary when i graduate
Hey i really appreciate your videos, way better explained than the average university prof. But your textbook link does not work. Thought id mention. Your website since Uvic also is not accessible for general public.
oh wow nvm clicked for 5th time and suddenly it works ¯\_(ツ)_/¯
What if I Taylored... (math analysis 1 flashback starts)
14th. I hope you are well.
my lord , thou art a savior .
5:20
i dont see why the integral of r'(x)/r(x) = ln(r(x))..
Let u=r(x). Then its du/u which intregrates to ln(u)
@@DrTrefor And where is dx?
Thank you, strange bearded Doctor!
💝
i dont have any money but have you considerd something like patreon? xD thank you so much
Dude... C=0 is *not* a recommended choice of C...
If you set C=0 for this setup, you'll just end up with a constant of 1 in front of the exponential. Since this constant will show up both above and below a division bar in the final answer, it doesn't matter what you make your +C. Because e^C will show up both above and below the division bar, and will cancel out of the equation.
There is an integral where it will matter, and where we'll ultimately solve for +C with an initial condition. But usually when there are more integrals in a solution, than there are orders of differentiation in the original equation, only as many arbitrary constants as the order of differentiation will need remain by end of your general solution, to solve for with initial conditions.
i love you
thats easy, i thought it would be harder since i am traumatised from the word "differential equations"...
sir is that a vape
no sense
didn’t understand shit
6:20