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!
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.
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!
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.
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
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.
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
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|
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
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.
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)?
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.
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
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.
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.
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
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?
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
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
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.
we cannot ignore the fact that he can flip pens instantly.
: )
That part!
you are a wonderful human
Uo o p p
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)
Hey Matthew! Hope all is well!
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!
I like how it looks like you're peeking from the right. I very much appreciate your explanation. Very helpful
can you make a tutorial of how to change markers with one hand quickly.
Thanks for saving me.
you're welcome
i was just so happy when i found out that you have a video about this theorem sir thank you so much !
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.
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!
thank you, textbooks could not explain this concept in human language, but you helped me for understanding what the "missing solution" is.
Love your math videos! makes me exited about maths again after chucking my text book out the window!
Thanku for this video.It helps me to understand the existence and uniqueness theorem in a proper way...May God bless u
I was so confused when my teacher taught this today, I love this man so darn much!😭😭! Thanks prof!❤
It might be part of the theroem, but I want to ask why we take the partial of y (df/dy), not x ?
Did you found an answer?
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.
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
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.
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
The Ivp dy/dx=3y^(2/3) ,y(0)=0 has two solutions or infinitely many solutions?
Please give reply sir
Why is x sqrt(y-3) continuous around (4,3)? It's undefined when y = 2.999.
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
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|
anas Nassar Without reading much into it, is that a "fancy" way of saying Continuous Around vs Continuous At?
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
Because of limits we re talking about continuity
My go-to math TH-camr
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
Love your pep and energy!!!
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.
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)?
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.
how did you determine at 1:41 that the function is continuous?
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
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.
THAAANKS! My exam is tomorrow and I had no idea how to solve this
Love your explanation and how you smile eachvtime,you make it look very exciting .keep on 🤗
Thank you for explaining this concept in detail.
thanks a lot, really clear
廖逸如 you're welcome 😊
May I know which book did you refer for this
That sleight of hand at 3:16 is probably why your channel is named blackpenredpen
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!
beastpenredpen coming in clutch yet again
appreciate this so much
Did you assume that Y cannot be 3 when you timed both sides by 1/sqrt(y-3)? But Y can be 3
I can’t help but laugh imagining a question going like
« Can you promise me there’s a unique solution? »
Nice explanation👌🏻
Love from india❤
you uploded this on my brithday
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.
You break it down!! Thanks.
At 2:17 why did you take partial with only respect y? Why did you differentiate with x as well?
possibly because it wasnt needed as taking a partial derivative and seeing it was niy continuous proves the question
Man, you save me in a differential equations
Thank you very much! You helped me clear a horrible confusion I had
really nice work, thanks!
y=3 is not missing solution, it is included when x=4
and x = - 4 also
Thaaaaaaaank You Maaaan ! God Bless You
So clear to understand, thank you sir!
Great video, thankyou. Hope you are well
thank you so much. your teaching is very engaging and clear.
Thank you so much 😭❤️❤️
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?
You are an amazing teacher!
thank God
Very good explanations, thanks
But how did you jump up to the solution of y=3? Quite obvious that it satisfies all the conditions but didn't undertsand.
Thank you professor.
Excellent explaining.
Sir Please, make conceptual video on this exact and uniqueness topic
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
mathphschjhb ((4)^2/4-4)^2+3=3
I understand why y=3 is a solution but is there an easy way to know to look for that?
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?
Thank you
Great explaining
you're a legend mate.
Thanks!
Thanks!
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
Thanks GOAT!
thank you very much.. you helped me a lot!!
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
would like to mention that I love your videos, though
Nice explanation ! Plz tell me that, is dy/dX=√|y| gives unique solution about (0,0). Plz reply me fast.
Thank you my lord
You sir are great.
Nyc explanation. So we can say that it possess only two solution Or we say that it has infinitely many solution? Plz explain sir
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
Thanks
thank you so much
Awesome video!
Can you please do some Cauchy problems with qualitative graph, Please help me
thank you sir
how do you find the missing solution if you are not given initial points?????
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?
Where dose the 2 in x/2(y-3)^1/2 come from?
its the derivative of it
Hey playlist doesn’t exists?
Great video. So neat. Thanks a lot! I really appreciate it :)
why is the partial derivative of x (y-3)^-.5 not (x(y-3)^.5)/2
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.
why only df/dy to check the uniqueness? How about df/dx? Is it enough only to check df/dy?
Great question
We can prove the solution isn’t unique as y=3 is a solution along with y=3+(1/4*x^2-4)^2
lovely!
why we have missing solution y=3, where it is come from ?
ivan luthfi y(4) = 3 thats where it comes from.
Thank you =)
you're welcome
When does uniquenes of ivp fails
that mic 😍😍😍
How will get the initial condition
Prasad Palle : )
Fantastic
Legend
is good lecturar
"does this have a unique solution? no. Can I still have one unique solution? yes." sigh...
Great
I am abangladeshi thanks