Existence & Uniqueness Theorem, Ex1

Existence & Uniqueness Theorem, Ex1
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[@siddheshss12345]

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

[@blackpenredpen]

: )

[@ShAlAmAnAyA3]

That part!

[@mtrichie111]

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)

[@blackpenredpen]

Hey Matthew! Hope all is well!

[@zackarygarner3396]

you are a wonderful human

[@ssjay7330]

Uo o p p

[@DawgFL]

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!

[@waltercarrillo6065]

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

[@beatrix2745]

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

[@Slapnutss]

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.

[@BOBPERIO2]

y=3 is an endpoint for f(x,y). Thus, f(x,y) is NOT continuous at the point (4,3). In other words, y-3 must be strictly greater than 0, NOT great than or equal to zero. This means we cannot even guarantee the existence of a solution to this differential equation. Good catch on the above video!

[@sakurastv]

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

[@babaika4941]

Did you found an answer?

[@nadiahmunirah6204]

Thanks for saving me.

[@blackpenredpen]

you're welcome

[@nicolasledoux1445]

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.

[@dimitriskayak]

i think because he wants to solve the ivp at x=4, he is not interested in taking restrictions, as 4^2/4 -4 =0 >= 0

[@thomaskim5394]

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.

[@NotYourAverageNothing]

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

[@blackpenredpen]

but y=2.999 is not in the domain of sqrt(y-3). We say a function is continuous if it is cont. at the values in its domain. so, sqrt(y-3) is cont. when y>=3

[@anasnassar8214]

Thank you for your good explanation. But as I know that for existence and uniqueness of solution, there should be a region surrounding the initial value y(x0)=y0 in which f(x,y) and df/dy are both continuous R={(x,y); |x-x0|

[@NotYourAverageNothing]

anas Nassar Without reading much into it, is that a "fancy" way of saying Continuous Around vs Continuous At?

[@anasnassar8214]

Yes, this is what I understand from the definition. the function must be continuous around this point (which also means that it's continuous at this point). Look at this notes (page 1 theorem 2): www.math.uiuc.edu/~tyson/existence.pdf

[@seoexperimentations6933]

Because of limits we re talking about continuity

[@eda7210]

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

[@tapaskumarpaul4982]

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

[@JakkalsJack]

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

[@michaelempeigne3519]

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

[@marylivesong357]

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

[@gracefully9699]

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

[@bharath6115]

Thank you for explaining this concept in detail.

[@eliseatkinson1832]

Love your pep and energy!!!

[@رغد-ظ5ف]

May I know which book did you refer for this

[@lilaitch705]

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

[@skylarculek]

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?

[@sripad72]

The condition that "partial derivative of f with y is bounded " is sufficient but NOT necessary. The condition fails does not guarantee that the equation has no unique solution. Please clarify.

[@yojacq]

you uploded this on my brithday

[@josephciesla536]

Thankyou, my professor is teaching with zoom and a horrible microphone and he is maddeningly boring, thankfully he has no due dates and I can just teach myself with HW and help from videos like these.

[@carylannebeltran]

Thank you!

[@ambrentu]

thanks a lot, really clear

[@blackpenredpen]

廖逸如 you're welcome 😊

[@thetedmang]

My go-to math KZbinr

[@irinaignatova1799]

I can’t help but laugh imagining a question going like « Can you promise me there’s a unique solution? »

[@strychnotic3933]

beastpenredpen coming in clutch yet again

[@Jhev1000]

This is a nice example to chew on. I'll be giving a lecture on the existence and uniqueness theorem in a few days. I'll probably steal this example!

[@teenaprajapati7678]

Nice explanation👌🏻 Love from india❤

[@craig8630]

appreciate this so much

[@ammarjaved3860]

That sleight of hand at 3:16 is probably why your channel is named blackpenredpen

[@kobodeb9912]

You're such a genius. Keep it up, but I would like to know how you were able know that y(x) = 3 is also a solution. We have come to know that it is a solution but my question rests on how we are able to know that. Could that proposition be based on any characteristic of the IVP? If so, could it be specified?

[@Keithfert490]

sqrt(y-3) is not continuous around y=3 as there is no positive constant, c, for which sqrt(y-3) exists when y=3-c

[@Keithfert490]

would like to mention that I love your videos, though

[@teja8575]

y=3 is not missing solution, it is included when x=4

[@michaelempeigne3519]

and x = - 4 also

[@dalal1047]

THAAANKS! My exam is tomorrow and I had no idea how to solve this

[@askog1900]

I understand why y=3 is a solution but is there an easy way to know to look for that?

[@Surendrarana-fn6pe]

Nice explanation ! Plz tell me that, is dy/dX=√|y| gives unique solution about (0,0). Plz reply me fast.

[@erikmjelde4428]

You break it down!! Thanks.

[@jojo23srb]

really nice work, thanks!

[@daninczech]

You're missing solutions because you're forgetting that the expression you get in the separation of variables holds only for y>3 and there are solutions obtained by gluing the two "types" of solutions you found

[@KingR787]

Great video, thankyou. Hope you are well

[@anaghgupta9348]

But how did you jump up to the solution of y=3? Quite obvious that it satisfies all the conditions but didn't undertsand.

[@benz8196]

when we say dy/dx= x* square root of(y-3), why in the second step @0:46 we say f(x,y) = dy/dx= x* square root of(y-3). Should we take a integral first before make it equal to f(x,y)?

[@jmac8561]

This is a year late but he's taking the derivative with respect to y so treat x as a constant. If you were using the product rule, it would have two functions with the same variable.

[@anilcelik7937]

how did you determine at 1:41 that the function is continuous?

[@anilcelik7937]

is it because we found a particular number by putting (4,3) to function? because we couldn't find any solution by putting (4,3) and we found it is discontinuous at 3:19

[@blackpenredpen]

Yea, it's easier to find that where the function is not continuous. If there's no place the function is not continuous, then its continuous. Usually check if the denominator is zero or not.

[@meenakshitripathi5159]

Nyc explanation. So we can say that it possess only two solution Or we say that it has infinitely many solution? Plz explain sir

[@lalaninishshanka8745]

thank you so much. your teaching is very engaging and clear.

[@user-od3mq7yd2x]

Thank you

[@nema1772]

Thank you so much 😭❤️❤️

[@mathphschjhb7749]

if (x^2 /4 - 4)^2 +3 is a solution, then it must satisfy the initial condition. I don't see how it does! even its derivative does not! pls explain... thank you

[@blackpenredpen]

mathphschjhb ((4)^2/4-4)^2+3=3

[@ShAlAmAnAyA3]

You are an amazing teacher!

[@nikitakipriyanov7260]

What is interesting this equation for each point y>3 has two different solutions. The special solution y=3 is an envelope of the set of all solutions with C≤0. It touches every solution such solution, and doesn't touch any of the solution where C>0: imgur.com/aB8odbb

[@benlybenthelegend9199]

Very good explanations, thanks

[@saurabhkujur6331]

Sir Please, make conceptual video on this exact and uniqueness topic

[@priklypaw2324]

Man, you save me in a differential equations

[@andrewrodriguez4516]

Thank you professor.

[@lal7030]

So clear to understand, thank you sir!

[@andrewcarter215]

Thanks!

[@RealLoveDragon]

Thanks GOAT!

[@katembogamesmrsf9498]

First : 5:28, you dont multiply the equation by dx, that is not correctly explained. Being a separable equation the member on the left becomes the derivative of a compound function and then you integrate on both sides, my teachers would give you 0 on that exercise because of that. Second: in the end you didnt exactly prove that there are multiple solutions, you have to draw the graphs and study it

[@helpfulfriend7762]

Thanks

[@sham5119]

thank you very much.. you helped me a lot!!

[@Fikayoz]

Thank you very much! You helped me clear a horrible confusion I had

[@cziegl3r]

Excellent explaining.

[@ralphyfabri4435]

Thaaaaaaaank You Maaaan ! God Bless You

[@MinuraPerera]

you're a legend mate.

[@blackpenredpen]

Thanks!

[@luvrajkhadkabk8715]

At 2:17 why did you take partial with only respect y? Why did you differentiate with x as well?

[@JensenPlaysMC]

possibly because it wasnt needed as taking a partial derivative and seeing it was niy continuous proves the question

[@mohammadghani2597]

I am still confused with your final conclusion that y(x) = 3 for all x, is that means all element x to be the solution?

[@prashanthreddyputta6611]

Can you please do some Cauchy problems with qualitative graph, Please help me

[@jafaraldahan4176]

Great explaining

[@evaristogabriel8208]

how do you find the missing solution if you are not given initial points?????

[@danzharzhavsky5461]

why is the partial derivative of x (y-3)^-.5 not (x(y-3)^.5)/2

[@proofofalifetime488]

Awesome video!

[@kanikagupta6103]

Hey playlist doesn’t exists?

[@kennedyada1117]

You sir are great.

[@averywlittle]

Where dose the 2 in x/2(y-3)^1/2 come from?

[@oinkvietnam4243]

its the derivative of it

[@melisaozen4897]

Great video. So neat. Thanks a lot! I really appreciate it :)

[@egeuludag972]

Thank you my lord

[@cameronspalding9792]

We can prove the solution isn’t unique as y=3 is a solution along with y=3+(1/4*x^2-4)^2

[@irfansyarafi800]

thank you sir

[@mohammadghani1379]

why only df/dy to check the uniqueness? How about df/dx? Is it enough only to check df/dy?

[@kobodeb9912]

Great question

[@yojacq]

thank God

[@vinsavi]

lovely!

[@jceepf]

At first it looks weird because dy/dy=x*f(y-3) where f(0)=0 guaranties that y(x)=3 is always a solution. So? We always have two? Well, of course not: when the existence and uniqueness theorem are satisfied, the constant solution emerges from the separation of variables! Funny video.

[@ivanluthfi8832]

why we have missing solution y=3, where it is come from ?

[@subscribes6434]

ivan luthfi y(4) = 3 thats where it comes from.

[@mallikarjuna4203]

When does uniquenes of ivp fails

[@ammelinecool]

Thank you =)

[@blackpenredpen]

you're welcome

[@yaseenali8318]

Legend

[@jakeyaboi6824]

Fantastic

[@prasadpalle2542]

How will get the initial condition

[@blackpenredpen]

Prasad Palle : )

[@alexsubba3565]

Great

[@jemalumer7333]

is good lecturar

[@terrancepage9163]

"does this have a unique solution? no. Can I still have one unique solution? yes." sigh...

[@prabhathkanishka3184]

that mic 😍😍😍

[@indrada-rf2vu]

Ha, funny enough when I solved the equation in my head I got y=0, which is always an annoying solution, but it was, a, solution

[@noahchung1944]

W video

[@shayobmathsolution5611]

I am abangladeshi thanks