How to Solve Clairaut's equation 😁

Подписаться 151 тыс.

Просмотров 6 тыс.

50% 1

🤩 Hello everyone, I'm very excited to bring you a new channel (aplusbi)
Enjoy...and thank you for your support!!! 🧡🥰🎉🥳🧡
/ @sybermathshorts
/ @aplusbi
en.wikipedia.o...
⭐ Join this channel to get access to perks:→ bit.ly/3cBgfR1
My merch → teespring.com/...
Follow me → / sybermath
Subscribe → www.youtube.co...
⭐ Suggest → forms.gle/A5bG...
If you need to post a picture of your solution or idea:
in...
#calculus #differentialequations
via @RU-vid @Apple @Desmos @NotabilityApp @googledocs @canva
An Exponential Equation | 2^{3^{5^x}} = 3^{2^{5^x}}: • An Exponential Equatio...
A Surprising Exponential Equation: • A Surprising Exponenti...
PLAYLISTS 🎵 :
Number Theory Problems: • Number Theory Problems
Challenging Math Problems: • Challenging Math Problems
Trigonometry Problems: • Trigonometry Problems
Diophantine Equations and Systems: • Diophantine Equations ...
Calculus: • Calculus

Опубликовано:

15 окт 2024

Ссылка:

Скачать:

Готовим ссылку...

Добавить в:

Мой плейлист

Посмотреть позже

Комментарии : 25

@michaelbaum6796 11 месяцев назад

Very nice ODE👍

@seanfraser3125 11 месяцев назад

As others have pointed out, you skipped the last step in both solutions, which is to plug them back into the original DE to check if they’re valid. Doing so for the first solution, we get kx+C = kx + e^k This means the solution is only valid for C=e^k. Thus the actual general linear solution is y=kx+e^k. k can still be any real number, and this solution is valid over all of R. For the second solution, using the fact that y’ = ln(-x), we get xln(-x)-x+C = xln(-x) + e^ln(-x) -> xln(-x)-x+C = xln(-x) -x So C=0, and we only get one solution y=xln(-x)-x. Obviously this is only defined for x

@SyberMath 11 месяцев назад

Good points!

@EugeneKogan-e8z 11 месяцев назад

Really liked the explanation!

@SyberMath 11 месяцев назад

Glad to hear that

@yoav613 11 месяцев назад

Nice,but the solution is not kx+c,but y=kx+e^k.and y=c is not asolution unless c=1.and also y= xln(-x)-x+c is not asolution unless c=0. Since you differenciated the original equation you should plug the solutions you got and see which constant c would give you the correct answer😃

@yoav613 11 месяцев назад

@@mrityunjaykumar4202 try to plug the solutions in the original solution,and you will understand.

@bobbyheffley4955 11 месяцев назад

This is the reason for checking solutions to make certain that they are not extraneous.

@jderick17 11 месяцев назад

It seemed interesting to me (I don't remember if mentioned in the video) the fact that the linear solutions of the equation are tangent lines to the other solution x(ln(-x) - 1).

@r2k314 11 месяцев назад

does this form of D.E. have any applications?

@dariosilva85 11 месяцев назад

Only the linear solution is valid. The other one doesnt check when you put it back in the original equation.

@yoav613 11 месяцев назад

And also not all linear solution works but y=kx+e^k.

@dariosilva85 11 месяцев назад

@@yoav613 Yes true.

@bobjazz2000 11 месяцев назад

Y=1

@dariosilva85 11 месяцев назад

@@bobjazz2000 That is not a new solution. It corresponds to the linear solution suggested above when k=0.

@cameronspalding9792 11 месяцев назад

@ 3:32 Shouldn’t you substitute into the original equation to see which constants work.

@SyberMath 11 месяцев назад

Yess

@arek10ful 11 месяцев назад

y=1

@mikeduffy4450 11 месяцев назад

You should have known your linear solution was incorrect because there is only one constant of integration, not two.

@tamilselvanrascal5956 11 месяцев назад

❤❤❤

@Kaan.- 11 месяцев назад

dude are we in same class bcs i was learning clairaut today.

@SyberMath 11 месяцев назад

No way! 😁

@giuseppemalaguti435 11 месяцев назад

x=t-2t/lnt...y=tlnt-t

@aashsyed1277 11 месяцев назад

1st!

@BridgeBum 11 месяцев назад

Call this a minor quibble, but I think the overlap of people who (a) can follow a complicated DE and (b) needs to have integration by parts explained is vanishingly small. I'll call it dIBP.