Dr. Bazzet! I've always been grateful to you🙏 As a math student, I have always been beyond amazed by the grandeur of mathematics.😲 It is PHENOMENAL! 💯💫
Generally don't leave comments but this video just cleared absolutely any and every bit of uncertainty I had with ODEs. Now I'm going to search for your videos on 2nd order ODEs!
This is a great explanation! My professor didn't explain this anywhere near this good and left me so confused but it's so much easier than I thought. Thank you!
Thank you so much sir. I was learning about the charging of a capacitor and got stuck at a linear first order ODE. I used this and the previous video to learn how to solve them, and I was finally able to derive the capacitor charging equation myself. Thanks a million from India!
first of your videos I've watched, I have an exam tomorrow and was just brushing up on a few niggly bits. This was extremely well explained, thank you!
This video made me angry. I just realized how poorly my professor explains the fundamentals of Differential Equations like integrating factors. I loved DE in undergrad and haven't really enjoyed it in the grad course. I hope your Fourier Series videos can save me. Thanks for the great videos!
If all universites had the teaching abilities you possess, the world would be filled with many fewer dissatisfied graduates, and enthusiastic engineers!
So if the product rule is an equality and a direct substitution for the left side of the equation, why do you still multiply the integrating factor by the right side? Because it is only a direct substitution after you multiply the IF by the whole equation?
If it is separable, then yes it is solvable always (up to our ability to do the integrals). However, the method typically will NOT be the method of integration factors.
One thing that is currently confusing me with this topic is why (e^(4x)(dy/dx)+4ye^(4x))=(d/dx)(ye^(4x)). every time i try to do it out myself i end up with it equaling (d/dx)(2y^(4x))
Greetings again from nyc! Another awesome video Dr. Trefor :) Got a quick question about some mathsss at the end. I paused the video and tried working the problem through first but when I tried to isolate y and divided the right hand side of the equation by e^4x i got y = (1/3e^x) + C/e^4x. I'm sure this id deliriously minor but I just don't understand how you got e^-4x on the right hand side...
Greetings! You and I are saying the same thing. That is, Ce^{-4x}=C/e^{4x}, this is the meaning of the negative in the exponent is basically that it puts the same thing with a positive down in the denominator.
@@DrTrefor AHHHH! Thank you so much! This was a case of-"i have been looking at these same numbers for so long that i feel loopy". Thank you for being so patient and clearing that up for me!
I don't know if anyone will read this but how could I have known that (e^4x)y'+(e^4x)4y is the equivalent to d/dx (e^4x)y? How was I supposed to come to the conclusion that it is. I know it works because of the product rule but that depends on me knowing the answer is (e^4x)y in the first place
You can verify it be working backwards with the product rule. It might be easier to see if you regroup it like this: The term that made it click for me was *(e^4x)4y* --> (4e^4x)y f(x) = e^(4x) f'(x) = 4e^(4x) g(x) = y g'(x) = y' so you can rewriting as: f(x) * g'(x) + f'(x) * g'(x) which is the same form as the product rule
A => would have been helpful on the solution to C line to disambiguate the equals chain. It was really confusing to many of my students. Something like this 4/3 = 1/3e^-(0) + Ce^(-4(0)) => 4/3 = 1/3 + C. I see this a lot actually, and I agree students should know these steps, but the students that didn't make A's-B's in lower classes may not all have this refined intuition.
these two videos on integrating factor may have been the 15 most educational minutes of my entire life, thank you Dr. Bazett
😢
Dr. Bazzet! I've always been grateful to you🙏
As a math student, I have always been beyond amazed by the grandeur of mathematics.😲
It is PHENOMENAL! 💯💫
Generally don't leave comments but this video just cleared absolutely any and every bit of uncertainty I had with ODEs. Now I'm going to search for your videos on 2nd order ODEs!
This is a great explanation! My professor didn't explain this anywhere near this good and left me so confused but it's so much easier than I thought. Thank you!
This channel is really underrated
Ha, I like to think so!
interestingly, i saw same topic yesterday in my lecture. He is still better than my instructors
Thank you so much sir. I was learning about the charging of a capacitor and got stuck at a linear first order ODE. I used this and the previous video to learn how to solve them, and I was finally able to derive the capacitor charging equation myself. Thanks a million from India!
first of your videos I've watched, I have an exam tomorrow and was just brushing up on a few niggly bits. This was extremely well explained, thank you!
Thank you. I needed this to refresh my memory.
Dr Bazett, thank you so much. The first video and this video really helped me understand Differential Equations.
Thanks for sharing this. It cleared up a couple issues I had solving some problems.
Such a clear, easy-to-follow explanation. Thank-you so much! :)
This video made me angry. I just realized how poorly my professor explains the fundamentals of Differential Equations like integrating factors. I loved DE in undergrad and haven't really enjoyed it in the grad course. I hope your Fourier Series videos can save me.
Thanks for the great videos!
Thanks for making this so easy to understand ❤️❤️❤️
If all universites had the teaching abilities you possess, the world would be filled with many fewer dissatisfied graduates, and enthusiastic engineers!
Professor Bazett, thank you for solving Linear First Order Ordinary Differential Equation using The Integrating Factors.
dude this content is gold
thank you so much
So if the product rule is an equality and a direct substitution for the left side of the equation, why do you still multiply the integrating factor by the right side? Because it is only a direct substitution after you multiply the IF by the whole equation?
Great explanation. Thanks, Doc!
this example saved me, literally
Awesome!! Dr.
Useful content that helped alot.
i appreciate it
Legend! Thanks!
Hello, I have a quick question. For can you solve a separable non-linear ODE's with methods we use for linear separable ODE's?
If it is separable, then yes it is solvable always (up to our ability to do the integrals). However, the method typically will NOT be the method of integration factors.
@@DrTrefor I see. Thank you for getting back!
Good video but kindly use highly contrasting colours for visibility, the background is soo dark
One thing that is currently confusing me with this topic is why (e^(4x)(dy/dx)+4ye^(4x))=(d/dx)(ye^(4x)). every time i try to do it out myself i end up with it equaling (d/dx)(2y^(4x))
What you said at 2:25 is not true if it's the derivative with respect to x. Could you explain, please?
right they dont cancel out they add together
Great vid mate keep up the good work 👍
Nicely explained
Thank you for this informational video.
Greetings again from nyc! Another awesome video Dr. Trefor :) Got a quick question about some mathsss at the end. I paused the video and tried working the problem through first but when I tried to isolate y and divided the right hand side of the equation by e^4x i got y = (1/3e^x) + C/e^4x. I'm sure this id deliriously minor but I just don't understand how you got e^-4x on the right hand side...
Greetings! You and I are saying the same thing. That is, Ce^{-4x}=C/e^{4x}, this is the meaning of the negative in the exponent is basically that it puts the same thing with a positive down in the denominator.
@@DrTrefor AHHHH! Thank you so much! This was a case of-"i have been looking at these same numbers for so long that i feel loopy". Thank you for being so patient and clearing that up for me!
I don't know if anyone will read this but how could I have known that (e^4x)y'+(e^4x)4y is the equivalent to d/dx (e^4x)y?
How was I supposed to come to the conclusion that it is. I know it works because of the product rule but that depends on me knowing the answer is (e^4x)y in the first place
You can verify it be working backwards with the product rule. It might be easier to see if you regroup it like this:
The term that made it click for me was *(e^4x)4y* --> (4e^4x)y
f(x) = e^(4x)
f'(x) = 4e^(4x)
g(x) = y
g'(x) = y'
so you can rewriting as: f(x) * g'(x) + f'(x) * g'(x) which is the same form as the product rule
Idk if you still answer these but at around 2:28 how did you simplify that.
essentially reverse the product rule
wow perfect DR
It was helpful.
Thank you.
Hi why you only use g(x) but not the f(x) ? Thank you in advance for any advice!
i still couldn't see it for the product rule. can you explain for me?
@ 0:22 Of course, 1 = y^0
Great video
Awesome, thanks a lot!
You're welcome!
You are the best
Why can't all maths professors be like this?
shouldn't the e^-(4x) be multiplied instead of added?
Yep, good thing he multiplied it.
God bless your soul
wouldn't it be c=-1 not c=1?
Thank you bro
amazing video
tysm for the 2 vids, 😀
thank you so much, my professor isn't bad but I just forget and can't find my notes
Dr please what is the best book of linear algebra if I want to learn more about it of course with your course🙂🙂
THis is an open source version: buzzard.ups.edu/
Thank you dr and thank you for your time you are great at all 😀😀😀
A => would have been helpful on the solution to C line to disambiguate the equals chain. It was really confusing to many of my students. Something like this 4/3 = 1/3e^-(0) + Ce^(-4(0)) => 4/3 = 1/3 + C. I see this a lot actually, and I agree students should know these steps, but the students that didn't make A's-B's in lower classes may not all have this refined intuition.
As a grade 9 student, I was able to comprehend that.
Why did 4y turn into just 4
Does anyone know what app he is using?
very nice
thanks boss
please blink for the love of god
Nice...
Thank you!
🔥🔥🔥
♥️♥️
HINDI
First : )
Nice one!!
@@DrTrefor Thank You Sir 😅,
You are very inspirational.
You teaching skill are really amazing ❤️
What happened to the Plus C in the integration factor