That was an excellent video, and very timely. I am compleyely self taught, and while I have a basic knowledge of ODEs and PDEs, I am really addicted to complex analysis, and tend to focus on it at the expense of other topics. Your detailed, explicit, and thorough coverage of this topic is just what I needed. So, that is a rather wordy way of saying "thank you"!
I like the video. There is a slight problem with the integration constant c. Although no set of numbers is specified, it seems to be implied that everything is supposed to be real. At 9:38 we obtain c as exp(c_3), which for real c_3 implies that c is positive. At 12:03 you state that c can have 'any value', without discussion. This is actually true, but cannot be tacitly implied from what went before. At 13:42 you actually discuss negative values of c. This means that we are apparently not relying on the context in which the ODE arose, e.g. rabbit population in Australia, which might require y(x)>0 everywhere. Then you really need to argue explicitly that c can be non-positive.
It is possible to be more rigorous in the procedure for solving the equation dy/dx = y. For example, one technically has to consider the possibility that y = 0. In this case, it is not possible to multiple both sides of the equation by 1/y, since 1/y does not exist. You would have to ask yourself, is it possible that y = 0? Well, if y = 0, then dy/dx = 0, and by transitivity, dy/dx = y, so yes, it is possible. You can set y = 0, and now restrict yourself to consider y ≠ 0, so now, it is meaningful to have (1/y)dy/dx = 1. When you antidifferentiate, though, you would need to be careful. For y < 0, the antiderivative of 1/y is ln(-y) + c0, while for y > 0, the antiderivative of 1/y is ln(y) + c1, and notice that c0 ≠ c1 is possible. Therefore, both of these cases must be accounted for, resulting in ln(-y(x)) + c0 = x + c2 for y < 0, and ln(y(x)) + c1 = x + c2 for y > 0. These can be rearranged to give y(x) = -(e^(c2 - c0))e^x and y(x) = (e^(c2 - c1))e^x respectively. We can simplify the results with the substitutions c3 = -e^(c2 - c0) and c4 = e^(c2 - c1). However, we need to be careful. e^(c2 - c0) > 0, regardless of whatever c0, c2 are, so c3 < 0 and c4 > 0. We have these restrictions for the possible values for c3, c4. Together with y(x) = 0, the equation is solved by y(x) = c3e^x, y(x) = 0, y(x) = c4e^x. Here, we can combine all three of the equations. Notice that 0 = 0e^x. Therefore, we can write y(x) = ce^x, where now, c can be any real number, and the three equations above are just the special cases where c < 0, c = 0, c > 0, respectively. This kind of rigor can be important when solving more complicated equations, where the constraints on the constants involved can actually be relevant.
Absolutely! We always have to make choices as to what to include in a shortist video, and our approach is somewhat pragmatic as we say in the intro with "maths for science and engineering". Very often the functions and equations of interest in science and engineering are "well-behaved", and this principle is guiding our choices. But we are actually preparing problems and solutions to go with the videos, and in those we do explore some of these subtleties in detail, including exactly what you've described. We hope to release the problems+solutions soon. Overall, thanks for the feedback!
That is fantastic news about problems and solutions -- they will be especially helpful in the more advanced videos undoubtedly to come!! Thank you!@@ProfessorMdoesScience
You are pure gem... You make things so easy to understand If possible may you please make some videos on advanced quantum mechanics( Perturbation theory, variational methods, scattering etc.) your videos are very helpful thank you very much ❤
Hello! I'm happy that you are Back, i was watching your videos, i Made a question in your time evolution operator video, i hope You can respond me profesor M, thank you !!
Brava. Clear, accessible -- without any loss of rigour. As always. Looking forward to this series -- and maybe some videos on Group Theory (tied back to your QM vids)? Thanks (so much!) for the channel.
@@snjy1619 Thanks for your answer! But still not sure I understand what you mean by "watch live"? That your professor played the video live in your lecture?
