Existence & Uniqueness Theorem, Ex1

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Existence & Uniqueness Theorem, Ex1

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we cannot ignore the fact that he can flip pens instantly.

siddheshss

2 Years after your class and I'm still using your videos to get through college, thanks a lot Profesor (Matthew K., Calc 2 Fall 2019)

mtrichie

This is a great video man. Simply and slowly explained like a highschooler could understand it. You're great at breaking it down to make it seem easy unlike so many other people on youtube, thanks!

DawgFL

can you make a tutorial of how to change markers with one hand quickly.

waltercarrillo

I like how it looks like you're peeking from the right. I very much appreciate your explanation. Very helpful

beatrix

In the first problem, how is f(x, y) continuous since left hand limit does not exist(limit y tends to 3-)( and consequently, is not equal to f(4, 3) and the right hand limit).Please explain sir! I'm very confused.

Slapnutss

It might be part of the theroem, but I want to ask why we take the partial of y (df/dy), not x ?

sakurastv

When you squared both sides, I think you should add the condition x^2/4-4 >= 0.Because when x^2/4-4 < 0, y = (x^2/4-4)^2+3 is not solution of the equation.We have 1/4*(-16+x^2)*x = -1/4*(-16+x^2)*x.

nicolasledoux

Notes integral of 1/√(y-3) exists if y is not = to 3 for the uncountable number of y values. The first answer is when y is not = to 3 for the uncomfortable number of x values. The second answer is when y=3 for the uncountable number of x values which is the case with the second answer.

thomaskim

Why is x sqrt(y-3) continuous around (4, 3)? It's undefined when y = 2.999.

NotYourAverageNothing

i was just so happy when i found out that you have a video about this theorem sir thank you so much !

eda

The Ivp dy/dx=3y^(2/3), y(0)=0 has two solutions or infinitely many solutions?
Please give reply sir

tapaskumarpaul

Love your math videos! makes me exited about maths again after chucking my text book out the window!

JakkalsJack

you say it is a missing solution, but it is covered by your differential solution when ( 1 / 4 )^x^2 - 4 = 0 which means that x = + or - 4

michaelempeigne

Love your explanation and how you smile eachvtime, you make it look very exciting .keep on 🤗

marylivesong

Thanku for this video.It helps me to understand the existence and uniqueness theorem in a proper way...May God bless u

gracefully

Thank you for explaining this concept in detail.

bharath

May I know which book did you refer for this

رغد-ظف

Did you assume that Y cannot be 3 when you timed both sides by 1/sqrt(y-3)? But Y can be 3

lilaitch

Is your function not discontinuous given that y < 3 returns a nonreal answer? In otherwords, it's continuous at 3 itself but not around 3, no?

skylarculek

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