ODE | Existence and uniqueness example

[@lisaswartz1475]

better in 4 minutes than in my teachers lecture 5 times that long

[@JoshBrownPhotography]

Easy voice to listen to, loud, clear, no heavy accent. Great illustration! Thank You! Please keep making more!

[@melvinguzman9353]

this man just bodied this lecture

8 years later, this video is still saving students

[@allpurposenerd4297]

Literally this 4:43 video taught me more on this theorem then a lecture! Thank you my friend.

[@happflapson]

It's really awesome that you made this video when I was a junior in high school, but it's helping me now when I'm a junior in college. Thanks!

[@khaledqaraman]

The best explanation I have ever seen on existence and uniqueness theorem Thanks a lot bro

[@jessicaparks3754]

Such a great visual of existence, uniqueness, and the two together! Thank you!

[@jeffreybrodhecker1367]

Dude thank you, my professor and textbook are absolute garbage. I couldn't figure out for the life of my what f(t,y) meant, when it's literally the simplest thing. Idk why switching around variables on absolute beginners is necessary but thank god I found this video

[@this_is_cricket]

Thank you so much for this. I’ve been having trouble getting through the existence and uniqueness part of a textbook and this helped me loads in getting past it.

[@parvizhasanzade271]

after 8years it is still useful

[@xyzsimplified1545]

Nice video. One comment: the exclusion of the vertical line as the solution curve passing through a point (0,b) is artificial and due to insisting on expressing solutions in the form y=y(x). If one rewrites the equation in a more symmetric form xdy=ydx, then the vertical line is a solution curve which is just as good as the horizontal line.

[@bryansalamat4505]

I think that: 1. The solution exist at a=0 (with any value of b, including b=0). However at a=0 and b=0 the solution is not unique. 2. The solution exist and is unique at a=0 provided that b is not equal to zero. In this case the solution is: x=0 (a vertical line along the y axis). To prove case 2, rewrite the equation to: dx/dy = x/y Applying the theorem, the solution exist and is unique provided that y is not equal to zero...

[@00purrington]

Well done! Easy to understand.

[@vaishnav4035]

Clear explanation 😃😃😃

[@MrExspectator]

Best explanation.

[@Kevin2341]

Thanks for the video, this makes a hell of a lot more sense than my book makes it (especially wikipedia)

[@Monaworld2023]

Very clear Interpretation,thanks!

[@ZomfgDaniel]

Great explanation. Thanks a lot!

[@JohnVKaravitis]

MIND BLOWN!

[@ilyaskhan6122]

find an interval centered about x = 0 for which the given initial-value problem has a unique solution. (x -2)y" +3y =x, y(0) =0,, y'(0) = 1

[@Fallkhar]

Why are we not taking the continuity (or lack thereof) of y, when considering the continuity of y/x?

[@victortopash950]

In the original differential equation, if x=0 then y=0. Does that bear any significance to the existence and uniqueness?

[@lolasogm]

hey man nice videos, do you have some on second order differential ecuations?

[@sandau44]

This is a very good video! Thanks.

[@muhammadsadiq1792]

good explanation

[@gvphysik]

Great content!

[@rubenlazell3003]

can someone help me? why is it when y= b/a * x at 3.30

[@sherevanalhamy9898]

You have a nice voice

[@howmathematicianscreatemat9226]

yeah, it's so badass motivational ;)

[@merameskul]

i appreciate that man, thanks a lot.

[@redreed42]

How about when b=0? Can someone please explain? Please and thank you!

[@pjg805]

b=0 would get you y(x)=((0)/a)*x = 0x =0 so then y=0. y=0 is just the x axis. y=0 is an equation obviously and is known as the trivial solution to a lot of diff equation problems. You can check that it works by plugging it in to the diff eq that he starts with: dy/dx = y/x = 0/x= 0 so dy/dx=0. Remember y=0 so that means y'=0, so 0=0. it works/checks out! :)

[@kendalwilliams5128]

this was great! thanks

[@istainblack]

I don't understand what the word "near " means in this context

[@surrendherify]

it means there exist 2 real numbers (let's call them c and d) that the function is continuous for every point inside the rectangle with width from a-c to a+c and height from b-d to b+d so basically, you can draw a small enough rectangle around the point (a,b) inside which the function is continuous at every point points inside that small enough rectangle can be called points "near" (a, b)

[@mohammadghani1379]

why only df/dy to make sure that solution UNIQUE? How about df/dx? Is it enough only df/dy that the solution UNIQUE?

[@mouthopenmonkey4571]

only variable y since that is the dependent variable. Also since the equation is dy/dx so you only need focus on y not x. If it was dx/dy, then u focus on x.

[@badpanda84]

?? so you dont even need to solve the DE to work out wheahter its unique

[@curtpiazza1688]

Great! 😊 🎉

[@hecz250]

Thank you!

[@xPloSiVgT]

Very nice

[@tamil575]

I think here, the choice of constant in wrong, because when in b/a , when a=0, the y tends to infinity and not passes through (0,0). apart from that really its good video

[@gabbyj9923]

Thank you

[@gabrielmalca1306]

THANK YOU

[@nathongknight]

Thank You!!!!!

[@randimafernando7215]

thanks man

[@rishabhpanday8023]

good

[@andrazfrelih8481]

you know nothing jon snow

[@hawina8548]

Thank you!