Existence & Uniqueness Theorem, Ex3.5

[@lennyatomz8389]

Thank you SO much for all of your help! I genuinely appreciate your mathematical expertise as well as your enthusiasm in how you share your insight! Keep up the great work , my friend. :-)

[@dikshakushwaha7081]

One thing I must say, We need teachers like YOU. Thankyou so so so Much for all your videos.(From India)

[@nedeninisorma1731]

ITS AMAZING MAN YOU MADE MY DAY YOU MADE ME LOVE THIS SUBJECT MORE THAN ANYTHING THANK YOU BROTHER

[@dipeshbarman2138]

Just wow. I have confusion, how and when to construct the solution function when existence and uniqueness theorem does not satisfied.But now I understand the story behind it.Thank you man

[@srg575]

This is very nice. Thank you for giving me insight.

[@anilcelik7937]

excellent example. im not saying only for this example, all of your examples about ivp are wonderful. thank you for your help!

[@tornikeonoprishvili5069]

thanks man. well explained.

[@hannakennedy3720]

thank you, i enjoyed your videos i will follow for all my future DQ struggles.

[@vialvile107]

@ 1:31 When you take the partial derivative, you do the derivative of 1/y, but is it not -1/y?

[@lonelyburger9069]

He is right.the derivative is 1/y square.....😂check again

[@afzolkarim6001]

But sq root of 25 is both +ve -ve 5.... plz reply

[@ghasemm7195]

You're a savior

[@SaifUlIslam-db1nu]

For those who want to quickly go through the steps, I wrote down the solution for looking through the steps quickly. Hope it helps! Does the following IVP necessarily have to have a unique solution? dx/dy = (x-1)/y, y(3) = -5 Sol: f(x, y) = (x-1)/y Putting in (3, -5) we see that our function is continuous. Thus, we now appyly partial differentiation. We get: df/dy = -(x-1)/y^2 Putting in (3, -5), our equation stays continuous. Thus, we can be sure that atleast one unique solution MUST exist as a 'solution'. Let us take our equation from the start: => dy/dx = (x-1)/y => y*dy = (x-1)*dx Integrating both sides, => INTEGRAL(y)dy = INTEGRAL(x-1)dx => y^2/2 = x^2/2 - x + C Multiply/Divide by 2, => y^2 = x^2 - 2x + 2C ( Or just C ) => y^2 = x^2 - 2x + C -----> EQ(A) Putting in (3, -5) => 25 = 9 - 6 + C => 25 - 3 = C => C = 22 Putting C in EQ(A), => y^2 = x^2 - 2x + 22 => y = + sqrt(x^2 - 2x + 22) ---- ANS(1) But this is not the right answer. If we put in x = 3, we get back y = 5, but y(3) = -5, so this answer may not be correct. Thus, the next (below) answer is correct. => y = - sqrt(x^2 - 2x + 22) ---- ANS(2)

[@dutoitvandyk2531]

how does Lipschitz conditions apply in this example?

[@Agus-of6rh]

Continuously differentiable implies locally Lipschitz.

[@vipinsaini5906]

Sir plz make proper playlist for IVP...so that we can easily find all of your videos

[@michelchaman6495]

Great video! I keep misreading your name as blackpendragon XD

[@marcushendriksen8415]

Fun problem, I solved it in the thumbnail before watching for the solution.

[@marcushendriksen8415]

I should add that this humble brag comes after watching countless of your other videos since this one. Once upon a time, I wouldn't have had the faintest idea of what to do with an equation like that - now it seems trivial! It's really amazing.