One solution you did not get is y=2ln|csch(-1/2 x +1)|. You made the assumption that the integral of 1/(u^2-1) is -tanh^(-1)(u), except this only works for u between -1 and 1. Outside this range, the integral is -coth^(-1)(u) which leads to a slightly different hyperbolic answer.
Yes, he needs to better explain that you can't just take a composition of a function and it's "inverse", unless it is an inverse on all it's domain which the trigonometric "inverses" certainly aren't.
I just realized that this is the equation for modeling pressure drag (i.e drag acting on a moving car, airplane etc that is caused by the flow separating thus creating a low pressure behind the object) except you multiply by air density, drag coefficient etc.
@@jongyon7192p For objects in free fall, u would be the velocity and c1 would be gravity (may be both signs depending on the coordinate system in use) . You may see that when plotted u converges to a value, which is actually the terminal velocity, which in turn is achieved when gravity and drag is in equilibrium.
don’t you have to do some division by square root of constant stuff for integral(1/x^2+c). That may be canceled out when undoing inverse functions, I haven’t checked myself, just thought of it.
Analyst perspective: You definitely described a cool family of functions with an interesting and unique property... but you were exclusive to global properties! You described functions which have the global property y'*y''=y'''. The family can be extended if you count local properties of functions; for example, functions which have vanishing 1st/2nd derivatives at certain points (like any critical point of a function, planar points on surfaces, etc).
for the pick 1 cases you can write a general solution by multiplying the answer by the sqrt of the original constant. Also, y could be any quadratic function since if you take the 2nd derivative of a quadratic you get 0 and obviously the 3rd derivative would also be 0 so you'd still get 0 = 0. Someone also already mentioned the coth thing
That wouldn't work -- if *f* is a quadratic, the second derivative is constant. That constant is zero if (and only if) *f* reduces to an (affine) linear function.
As far as it seems, the other cases are the same. In the case of x-c, you make c(x/c-1) and making a few arrangements you come to the same, and for the case of x+c, it's analogous
Your Maths videos are very interesting dude. I am currently in my first year studying Engineering at Cambridge University and I watch your maths videos as they involve some unique problems such as this video right here! xD
AndDiracisHisProphet Oh yea I know. However I did think about how to really come up with the general form. But again, I don’t think it’s pretty. So yea.
Cool calculation. It would be fascinating to see a graphical/geometric interpretation. A functions whose slope multiple by its inflection equals its ....? Or perhaps include the z dimension somehow.
5:19 XAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXA Note: X in Russian is H in English.
Even though the method you used in the previous video was too long for this kind of problem, is there any case where that approach can actually be used? Can you make a video of such a case?
there is a mistake it's not |-u²| = u² , it's |-1|*|u²| = |u²| = |1| = 1 but u doesn't includ in reel number because u²+1 =0 have no solution in reel so |u²| different of u². ;)
I don't think it is. There will be 2 A's on the left side and the exponents will add up to make 2b. I'm just saying this by inspection so I might be wrong.
If we rewrite the case where c20, then we have (1/2a)[du/(u+a)-1/(u-a)]=(1/2)dx. Integrating gives (1/2a)Log[Abs[(u+a)/(u-a)]]=(1/2)x + b/2, where b is a constant. This gives Log[Abs[(u+a)/(u-a)]]=ax + b. Exponentiating both sides gives Abs[(u+a)/(u-a)]]=Exp[ax+b]. Solving for u gives u = a*[(Exp[ax+b]+1)/(Exp[ax+b]-1)]. Multiplying numerator and denominator by Exp[-b] gives u = a[(Exp[ax]+Exp[-b])/(Exp[ax]-Exp[-b])] = y'. Integrating gives the general solution for the case c2 Infinity gives the linear solution y = ax + c.
I know it’s the same thing, but I really hate the ln(sec..) solution... why not just use ln(cos..)? I actually hate all sec, csc, cosh, sinh.. notations and prefer to use 1/cos,(e^x + e^-x)/2.. Wonderful job as always. Also, if you decide to eliminate all linear solutions (so that y’’ isn’t zero), you could say y’=y’’’/y’’ so y=c+ lny’... no idea how to go from here but that was my second instinct.
My first thought when I saw the title was that you were either mupltiplying the complex conjugate of y' by y'' OR that you were convolving y' with y''. I completely forgot that normal multiplication can use an asterisk. I'm a little broken at this point.
13:34 = why did you say don't worry it is possible with the ax + bi scenario? Or am I missing something fundamental that should have required me to pay more attention in class 11.
really u wonderful zakeeri lee..and intersting..thank u for u efforts..i very interst about u explanation, but be more slow in ur talking for more interest
As n goes to infinity it gets bigger than any x, so (x-n)! isn't defined. Unless x doesn't assume integer values, then you could use the gamma function, but it would be crazy.
It is relatively straightforward to integrate in the case that c2 > 0? The indefinite integral would come out to atan(x/sqrt(c2))/sqrt(c2) + const - is there a specific reason we are choosing c2=1?
Just to be certain, the term that sounds like chenloo is actually "chain rule," right? The automated captions seem confused as well, over and over and over. (Chen Lu and chendu are my favorite caption guesses.)
1. Why didn't you do general cases for C =/= 0? They are both relatively simple from a table of integrals. 2. Don't forget that any linear combination of those solutions is a solution 3. Why can you get 4 solutions for an 3rd order differential equation, I thought the maximum was equal to the order?
1) True, I assume he was tired and found it trivial enough to leave it to us. 2) Assume y = f+g where f and g are solutions to y' y" = y''' y' y" = (f' + g')(f" + g") = f'f" + g'g" + f'g" + f"g' = f''' + g''' + (f'g')' this is not necessarily equal to f''' + g'''
Old post, but do you mind explaining what all the C's mean? I know you said they were constants, but what's their value, purpose, and relationship with one another?
the derivative of f(x)+c is the same for all possible constants c. for example the derivative of x² + 5 is the same as the derivative of x² + 10 with the integral he is undoing a derivative. but the solution is only unique except of the constant. if you have some number multiplied by another number you get some other number. So instead of writing 2*c1 he just writes c2
c is arbitrary, which means that it can be any constant. Because differentiating a constant gives you 0 y = c (c is constant) is just a straight horizontal line. Since differentiation is used to find the equation of the gradient at any point, throughout the horizontal line, the gradient is 0 (flat line) so differentiated constant is 0. Since it can be any constant (dy/dx 92937 is the same as dy/dx 1 because its just a straight line at different y values) c is arbitrary and can be any constant. The subscript for the numbers are just there to imply that it is a different random number to the previous (because it was multiplied/added to/divided etc)
Ok so I saw this and was wondering why that was true since I'm currently in Geometry And the moment you said "derivative" I knew id have to wait a couple years
Shop an Euler's Identity t-shirt: amzn.to/427Seae
Imagine randomly walking into the classroom and spotting a guy talking to a black ball.
also talking and watching onto laptop
While saying 'i dont like to be on the bottom'
You probably like horror game such as "Detention"
I would immediately know it is a goa'uld spy.
And then at 22:00...
One solution you did not get is y=2ln|csch(-1/2 x +1)|. You made the assumption that the integral of 1/(u^2-1) is -tanh^(-1)(u), except this only works for u between -1 and 1. Outside this range, the integral is -coth^(-1)(u) which leads to a slightly different hyperbolic answer.
I think bprp did a video on this "consider the domain" trap, too
Yes, he needs to better explain that you can't just take a composition of a function and it's "inverse", unless it is an inverse on all it's domain which the trigonometric "inverses" certainly aren't.
please stay healthy! don't overstress
we want you stay active for long, not for many content at once
NoName awwww thank you!!
It's up to "u".
y?
@@bobajaj4224 because we want to isolate "u"
@@leekon_ lol
5 points to griffindor
Yes, right never forget about Cheng Lu.
:')
Did I miss any solution?
If so, then you go find it.
What if C2 is imaginary lol?
y(x)=exp(cx) with "c" constant is also a solution for y'y''=y'''
@@vitorlazaroti4007 Nope. Left contains C^3 * exp(2x) and right contains C^3 * exp(x)
07:42 just sub u=sqrt(c_2)w...
General answer: y = -2 log(cos((sqrt(c_1) (c_2 + x))/sqrt(2))) + c_3
c_i \in C, i \in [3]
@@mauricereichert2804 Yep, thank you for that conclusion. Next time keep in mind we all who are interested in such problem understand this sub ;))
I just realized that this is the equation for modeling pressure drag (i.e drag acting on a moving car, airplane etc that is caused by the flow separating thus creating a low pressure behind the object) except you multiply by air density, drag coefficient etc.
where can i learn more about this? is the result the same as this video? graphing c
@@jongyon7192p For objects in free fall, u would be the velocity and c1 would be gravity (may be both signs depending on the coordinate system in use) . You may see that when plotted u converges to a value, which is actually the terminal velocity, which in turn is achieved when gravity and drag is in equilibrium.
I'm really sorry I made you do all this work ☺ (please do general case)
It's okay. It was a really good one and I add more primes to it for the extra fun. Very satisfying!
y=2*ln|sec(Ax + B)| + C for C2 > 0.
y=2*ln|sech(Ax + B)| + C for C2 < 0.
Worked it out for fun ;)
@@karolakkolo123 it wasn’t hard, was that?
don’t you have to do some division by square root of constant stuff for integral(1/x^2+c). That may be canceled out when undoing inverse functions, I haven’t checked myself, just thought of it.
Very nice indeed !!
Already checked in the comments, but you can generalize both C2 >0 and
Really love your dedication man, keep up the good work❤
HasBrah thank you!!!
I found another solution
-2/x * 2/x^2 = -4/x^3
Obtained from solving
y = ax^r
ax^r * arx^(r-1) = ar(r-1)x^(r-2)
You’ve found that y’ = -2/x which is covered by his first case since it gives y = -2 ln|x|
Analyst perspective: You definitely described a cool family of functions with an interesting and unique property... but you were exclusive to global properties! You described functions which have the global property y'*y''=y'''.
The family can be extended if you count local properties of functions; for example, functions which have vanishing 1st/2nd derivatives at certain points (like any critical point of a function, planar points on surfaces, etc).
@@akashthiagarajan4751 We'll all make it, chief
The first part of this process was so clever, loved that!
Any a,b for y=ax+b
You will get 0 = 0 for both sides of equation
Filming this awesome video late and while tired just because your last one "wasn't good enough". We don't deserve you blackpenredpen.
Very interesting and helpful videos, i actually began to understand maths by watching your videos. Keep it up man!!!
I love how you say the hyperbolic sine and cosine function as "sinch over cosh"
jake & cosh
Definitely worth a thumbs up. Love your videos!
berenjervin thank you!!!!
for the pick 1 cases you can write a general solution by multiplying the answer by the sqrt of the original constant. Also, y could be any quadratic function since if you take the 2nd derivative of a quadratic you get 0 and obviously the 3rd derivative would also be 0 so you'd still get 0 = 0. Someone also already mentioned the coth thing
That wouldn't work -- if *f* is a quadratic, the second derivative is constant. That constant is zero if (and only if) *f* reduces to an (affine) linear function.
@@carstenmeyer7786 I realized that as soon as I made the comment lmao
As far as it seems, the other cases are the same. In the case of x-c, you make c(x/c-1) and making a few arrangements you come to the same, and for the case of x+c, it's analogous
One of your best videos. Absolutely love it!!
This man really makes me learn math and enjoy it, amazing!
So late still doing math .... respect :) .... will we ever see a blooper compilation?
Л.С. Мото maybe?! Lol
Lol, I didn't know I needed one until you mentioned it
This is literally the best RUclips channel
These references to other videos make me laugh every time. Great Video as always, keep up the good work.
Do the general case!
It's actually just y=2*ln|sec(Ax + B)| + C. I worked it out on paper.
i did not know about the method to set u=x' and make f(x)dx=g(u)du.
now i have a new weapon to solve some differentiation problems. thanks!
Oh weeell, this was an awesome video! 👍
Your Maths videos are very interesting dude. I am currently in my first year studying Engineering at Cambridge University and I watch your maths videos as they involve some unique problems such as this video right here! xD
You dont need to put in diffrent cases, you can integrate int(1/(x^2+c))
How?
u^2+c=c(u^2/c+1)
int(1/(u^2+c)) = 1/c*int(1/(u^2/c+1))
t=root(x^2/c)) => 1/c*int(root(c)/(t^2+1))
=root(c)/c*int(1/(t^2+1))=root(c)/c*arctan(t)
finally this equals to root(c)/c*arctan(root(u^2/c))
@@blackpenredpen Is that solution valid?
"Ask and you shall receive!"
I really appreciate you doing the work that I just could not bring myself to do because it was after midnight.
👍👍👍👍👍👍👍👍👍
Seriously... THANKS!
Amazing!!! 😍
I cant even begin to comprehend this. Very intimidating proof
Differenzialgleichung. Unironisch mein liebstes Mathematikthema.
Can‘t proof it, but you did a fabulous, interesting and exhausting work!
(Do appreciate the heart...)
Don't worry, you're not the only one up late, watching this at 12:20 AM
7:41 please do general case.
AndDiracisHisProphet maybe!
@@blackpenredpen admit it. you just wanted us to beg you^^
AndDiracisHisProphet I actually don’t know how to make the general case in a pretty form. I think I will just c2 as k^2 and -m^2 and continue?
@@blackpenredpenthat's what i thought. I don't know if it is pretty or not, though.
also, of course i was joking. just to make it clear :)
AndDiracisHisProphet
Oh yea I know. However I did think about how to really come up with the general form. But again, I don’t think it’s pretty. So yea.
redpenbluepen: *puts integral sign before equal sign*
literally everyone in my class: *write integral sign upside-down*
Cool calculation. It would be fascinating to see a graphical/geometric interpretation. A functions whose slope multiple by its inflection equals its ....? Or perhaps include the z dimension somehow.
These videos are excellent and very helpful
Thanks a lot for your videos, I love them !!
5:19
XAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXAXA
Note: X in Russian is H in English.
Very clever seeing the chain rule
It took a while until I realized that 'tjin doo' is 'chain rule' 🙂
Why oh why would someone ask why a Y plus another Y equals more Y's?
General case was a factoring and a substitution away, although I do feel like I wouldn't do it either at 10:30 after class lol, nice vid
12:31
love this laughter
and me too 😅😅
my man's taught me stuff teachers couldn't for years
Wait till you find out he's a teacher
@@AmogUwUs my point you
Even though the method you used in the previous video was too long for this kind of problem, is there any case where that approach can actually be used?
Can you make a video of such a case?
for cases 2 and 3: if you make the substitution u^2 / c2 = f, you can always reduce to the cases c2 = 1 or -1 right?
You really did this at night and no one is with you? That's awesome!
Yup
@@blackpenredpen I'm noticed 😮😮😮
I like the way BPRP says 'thirteen'
Awesome! Please explain the elyptic integrals! Thank you
13:31 why you can't try it
u²+1=0
u²-(-1)=0
-u²-1=0 (dividing both sides by -1)
-u² = 1
|-u²|=|1| (taking modulus on both sides)
u²=1
u=1,-1
there is a mistake it's not |-u²| = u² , it's |-1|*|u²| = |u²| = |1| = 1 but u doesn't includ in reel number because u²+1 =0 have no solution in reel so |u²| different of u². ;)
31 years old, friday night, half drunk.... why am I watching this?.. oh no, why did I watch it till the end?
10:05 Last time you wrote inverse tanh backwards, this time you just point :)
fantastic solution🙏🙏🙏
Bprp: And as always, that's ALL
Me: Wait. That's illegal.
Bprp: I mean, that's it
oof
Hes going to math jail.
And as always...
THX FOR WATCHING! :3
This stuff is always so impressive to me
Ae^bx +C is one of the perfect solutions
I don't think it is. There will be 2 A's on the left side and the exponents will add up to make 2b. I'm just saying this by inspection so I might be wrong.
Special cases, for case 1, how can u^2 be zero? That will make a denominator zero.
If we rewrite the case where c20, then we have (1/2a)[du/(u+a)-1/(u-a)]=(1/2)dx. Integrating gives (1/2a)Log[Abs[(u+a)/(u-a)]]=(1/2)x + b/2, where b is a constant. This gives Log[Abs[(u+a)/(u-a)]]=ax + b. Exponentiating both sides gives Abs[(u+a)/(u-a)]]=Exp[ax+b]. Solving for u gives u = a*[(Exp[ax+b]+1)/(Exp[ax+b]-1)]. Multiplying numerator and denominator by Exp[-b] gives u = a[(Exp[ax]+Exp[-b])/(Exp[ax]-Exp[-b])] = y'. Integrating gives the general solution for the case c2 Infinity gives the linear solution y = ax + c.
y=a and y=ax+b are also solutions.
Can you do a general case
I know it’s the same thing, but I really hate the ln(sec..) solution... why not just use ln(cos..)? I actually hate all sec, csc, cosh, sinh.. notations and prefer to use 1/cos,(e^x + e^-x)/2..
Wonderful job as always.
Also, if you decide to eliminate all linear solutions (so that y’’ isn’t zero), you could say y’=y’’’/y’’ so y=c+ lny’... no idea how to go from here but that was my second instinct.
Awesome video, really enjoyed it :) Now I hope you get some rest :D
t. gobold I did. Thank you!!!
But where is c8?
Just kidding. I love your videos and your devotion at teaching, whether it's Maths or having a big smile after a long day!!
Dimitris Dimitriadhs I used it but changed it to c9 after multiply the negative
@@blackpenredpen I didn't realise it. Thanks for your reply.
8:40 the integral of the tangent isnt -ln(cos(x))???
sec(x)= 1/cos(x) so it's the same
I've immediately come with an exponential solution y=exp(ax+b)
OH, I thought we were going to see the final solution of the integral of the equal sign :c
My first thought when I saw the title was that you were either mupltiplying the complex conjugate of y' by y'' OR that you were convolving y' with y''. I completely forgot that normal multiplication can use an asterisk. I'm a little broken at this point.
Excellent huge treat
This deserves more than a thumb up. Congrats!
Fux Premier thank you!!
I love the way he says "chain rule"
5:08 not sure that you can do that because it's a Ricatti equation
13:34 = why did you say don't worry it is possible with the ax + bi scenario? Or am I missing something fundamental that should have required me to pay more attention in class 11.
The first few times I thought you were saying “reverse Chen Lu” instead of chain rule
thank you little dificolte and incomprehensif
really u wonderful zakeeri lee..and intersting..thank u for u efforts..i very interst about u explanation, but be more slow in ur talking for more interest
Bruh.... My mind's expanding
Thanks!
Is it possible to find the limit of (x-n)(x!)/(x-n)! an n goes to infinity?
As n goes to infinity it gets bigger than any x, so (x-n)! isn't defined. Unless x doesn't assume integer values, then you could use the gamma function, but it would be crazy.
Great video
It is relatively straightforward to integrate in the case that c2 > 0? The indefinite integral would come out to atan(x/sqrt(c2))/sqrt(c2) + const - is there a specific reason we are choosing c2=1?
DO THE GENERAL CASE!
It's actually just y=2*ln|sec(Ax + B)| + C. I worked it out on paper.
Just to be certain, the term that sounds like chenloo is actually "chain rule," right? The automated captions seem confused as well, over and over and over. (Chen Lu and chendu are my favorite caption guesses.)
Yes you are correct
1. Why didn't you do general cases for C =/= 0? They are both relatively simple from a table of integrals.
2. Don't forget that any linear combination of those solutions is a solution
3. Why can you get 4 solutions for an 3rd order differential equation, I thought the maximum was equal to the order?
1) True, I assume he was tired and found it trivial enough to leave it to us.
2) Assume y = f+g where f and g are solutions to y' y" = y'''
y' y"
= (f' + g')(f" + g")
= f'f" + g'g" + f'g" + f"g'
= f''' + g''' + (f'g')'
this is not necessarily equal to f''' + g'''
Awesome video !
Can you try to find a function f with :
f(x^2) + 2 = f(x).f(x) ?
Good luck :D
If someone find the solution do not hesitate to comment it pls
I immediately thought "y=c". But I figured there would be more.
how would you go about proving the general cases?
Old post, but do you mind explaining what all the C's mean? I know you said they were constants, but what's their value, purpose, and relationship with one another?
the derivative of f(x)+c is the same for all possible constants c.
for example the derivative of x² + 5 is the same as the derivative of x² + 10
with the integral he is undoing a derivative. but the solution is only unique except of the constant.
if you have some number multiplied by another number you get some other number. So instead of writing 2*c1 he just writes c2
c is arbitrary, which means that it can be any constant. Because differentiating a constant gives you 0
y = c (c is constant) is just a straight horizontal line. Since differentiation is used to find the equation of the gradient at any point, throughout the horizontal line, the gradient is 0 (flat line) so differentiated constant is 0.
Since it can be any constant (dy/dx 92937 is the same as dy/dx 1 because its just a straight line at different y values) c is arbitrary and can be any constant. The subscript for the numbers are just there to imply that it is a different random number to the previous (because it was multiplied/added to/divided etc)
Thanks for the your cours
Ok so I saw this and was wondering why that was true since I'm currently in Geometry
And the moment you said "derivative" I knew id have to wait a couple years
3:05 how you know what the integral to u•u' ?( you are writing dX ... this need ti be dU)
You can think of u' as du/dx. Substitute that in and the dx's cancel out giving you u•du.
@@badrunna-im great!!
If y=e^kx
y'*y''=y''', right?
Sir, make some videos on tensor analysis, it's too tough to grasp tensor. Please sir
Go to Leonard Susskinds lectures on General Relativity. He does an excellent introduction on Tensor Analysis
Excellent ! Integrate sqrt(cosx) I want to see this.
uhhh, no you don't
www.wolframalpha.com/input/?i=Integrate+sqrt(cosx)
it's a weird elliptical integral
Goddamit
In 5:09 how did you just integrate ?
10:12 no not okay