Abel's brilliant trick for solving differential equations

Рет қаралды 15,796

Maths 505

Күн бұрын

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@orionspur 9 ай бұрын

Totally guessed sin(x)arccoth(sin(x))-1 out of thin air while both cooking and mowing my lawn.

@maths_505 9 ай бұрын

Average maths 505 enjoyer 😎😎😂😂

@kevinmadden1645 9 ай бұрын

Life is a bummer without a good sense of humor.

@edmundwoolliams1240 9 ай бұрын

Could you pop-in a u-sub: u=sin(x) and have solved it?

@sametyetimoglu6026 9 ай бұрын

Hello? Ramanujan?

@CM63_France 9 ай бұрын

Hi, "ok, cool" : 3:42 , 7:12 , 7:40 , 9:18 . "terribly sorry about that" : 5:24 .

@Dedicate25 9 ай бұрын

my fav integral solver ❤

@qwertyman123 9 ай бұрын

i see you on cipher unity

@Dedicate25 9 ай бұрын

@@qwertyman123 :)

@qwertyman123 9 ай бұрын

@@Dedicate25 good to see another person who truly enjoys math

@mcalkis5771 9 ай бұрын

What a brilliant method. Thank you for showing this

@aliaujla-2ujt 9 ай бұрын

Bro please make a video in detail about cauchy residue theorem.

@randomzhjioewmx 9 ай бұрын

Here is another (essentially the same) method to obtain y_2 knowing already y_1=sin(x). We aim to factor the operator D^2-tan(x)D+2. This operator vanishes at sin(x), which also is a zero of the operator D-cot(x), so one may use the ansatz D^2-tan(x)D+2=(D-f)(D-cot(x)). Multiplying out and using Dcot(x)=cot(x)D-1/sin^2(x), one obtains D^2-tan(x)D+2=D^2-(f+cot(x))D+fcot(x)+1/sin^2(x). Comparing coefficients, tan(x)=f+cot(x) so f=tan(x)-cot(x). [One may check that this satisfies 2=fcot(x)+1/sin^2(x) so the factorization works]. So to find another zero h(x) of the operator (D-f)(D-cot(x)), we can first find a zero g(x) of D-f, and then solve (D-cot(x))h=g. To find g, one must solve g'-fg=0, so g'/g=f, so ln(g)=integral(f)=-log(sin(x)cos(x)) [we wlog choose 0 for the constant of integration) so g=1/(sin(x)cos(x)). To find h, we must solve the linear equation h'-cot(x)h=g. We make ansatz h(x)=sin(x)m(x). Thus sin(x)m'(x)=g, so m'(x)=1/(sin^2(x)cos(x)). Integration yields m(x)=arctanh(sin(x))-1/sin(x)+c. Thus h(x)=sin(x)arctanh(sin(x))-1+c sin(x). So one may set y_2=sin(x)arctanh(sin(x))-1.

@davidblauyoutube 9 ай бұрын

Amazing solution development!

@gideonbrown5093 9 ай бұрын

Wow reducing the second order to the first order using any arbitrary w where w is the wronskian is super cool…😍🥰🥰🥰😍😍

@bandishrupnath3721 9 ай бұрын

Sir pls make some videos which discusses different types of things like melin transform,laplace transformation as some of ur viewers are illiterate(Me also especially) compared to ur knowledge in calculus 🙃

@kappasphere 9 ай бұрын

Laplace transforms are really fun even though I still feel they're above my math skill

@alexander_elektronik 9 ай бұрын

i think you forgot to multiply the C_2 with y_1 at 9:33 but since you set c_2 = 0 it wouldn‘t make a difference :)

@aravindakannank.s. 9 ай бұрын

wow what a way🤯

@jieyuenlee1758 9 ай бұрын

9:56 Int(secx/sin²x)dx =Int(secxcsc²x)dx =-secxcotxdx+int(cotxtanxsecxdx) =-1/(tanxcosx)+ln(secx+tanx) =-1/sinx+ln(secx+tanx) =ln(secx+tanx)-cscx =ln((1+sinx)/cosx)-cscx =1/2ln[(1+sinx)²/cos²x]-cscx =1/2ln[(1+sinx)²/(1-sin²x)]-cscx =1/2ln[(1+sinx)/(1-sinx)]-cscx

@usmansaleem3173 9 ай бұрын

Kindly mention white board software you are using is also the name of the pen tablet

@dharunpranay8581 9 ай бұрын

I am feeling ill now for studying engineering .But I love maths the most than anything. But our India is not conducting such competitions like mit this makes me 😢

@AdrianRif 3 ай бұрын

Very elegant solution, but I think to make the equation more general you should have multiplied the hyperbolic cotangent term inside the parenthesis by the constant B as well!

@farfa2937 9 ай бұрын

Since x only shows as sinx, would it be correct to ditch the sin and change the domain to [-1,1]?

@fartoxedm5638 9 ай бұрын

I can't sleep well when B's at the end disappear... Absolutely fascinating technique tho! Thanks for sharing

@gerryiles3925 9 ай бұрын

In your final answer, the coth-1 should have been multiplied by B...

@maths_505 9 ай бұрын

Oh right my bad

@pandavroomvroom 9 ай бұрын

"your fav integrator on youtube" - well its true

@sgiri2012 9 ай бұрын

Sir, can this differential equation gives answer when we try the series solution or frobenius solution depending upon the case?

@maths_505 9 ай бұрын

Yup

@Mr_Mundee 9 ай бұрын

do this : integral from 0 to infinity of tanh(x)ln(coth(x))dx

@maddog5597 9 ай бұрын

Am I missing something? The arccoth(x) is not real for abs(x)< 1, which is pretty much where the sin(x) lives. If you want to include complex solutions, fine. But I have a feeling we were dealing with real quantities.

@randomzhjioewmx 9 ай бұрын

Replace arccoth by arctanh

@maddog5597 9 ай бұрын

@@randomzhjioewmx Um, OK, but why? The solution as presented is arccoth. Unless that’s wrong. He did manage to ignore a sign along the way.

@randomzhjioewmx 9 ай бұрын

@@maddog5597 The correct solution is indeed with arctanh instead of arccoth. In the video, the arccoth appears in the claimed identity 1/2 log((1+sinx)/(1-sinx))=arccoth(sinx) at 11:16, but this is incorrect and one must use arctanh here. This is because one has 1/2 log((1+y)/(1-y))=arctanh(y) if |y|

@maddog5597 9 ай бұрын

@@randomzhjioewmx Thanks very much for the clarification. I really do wish Maths505 would use a script and rehearse these postings. He makes so many errors like this…

@alecbg919 9 ай бұрын

This channel is awesome. 10:20 should be C_2 sin(x) right?

@maths_505 9 ай бұрын

Thanks Yeah it should be C_2 sin(x). Didn't matter as C_2 is zero anyway

@RobMartin-gz3zk 8 ай бұрын

5:29 so close

@거미남자_spidy 6 ай бұрын

divide...each term by y' ??

@zygoloid 9 ай бұрын

Incredibly, this doesn't seem to make obvious use of *any* special properties of W. I wonder if other functions combining y1 and y2, instead of W, would also work.

@pokakoka4099 9 ай бұрын

Brilliant sir, sir just tell me the name of app you are using for video recordings. Will wait sir

@maths_505 9 ай бұрын

Samsung notes

@maths_505 9 ай бұрын

And the screen recorder

@pokakoka4099 9 ай бұрын

Thanks alot sir

@balubaluhehe2002 9 ай бұрын

Hey! Are you going to solve the 2024 MIT integration bee semi-final and quarter-final problems? I saw some quite interesting ones there

@maths_505 9 ай бұрын

Yeah I plan to solve a few problems from them too

@balubaluhehe2002 9 ай бұрын

@@maths_505i'll be waiting for them

@edmundwoolliams1240 9 ай бұрын

Nice method! What would you do if that 2y in the original ODE was 3y?

@maths_505 9 ай бұрын

Probably a drastically different solution development 😂

@edmundwoolliams1240 9 ай бұрын

😂 Could you pop in a series solution in that case? Or would the tan just make that intractable?

@maths_505 9 ай бұрын

@@edmundwoolliams1240 a series solution does seem promising even with the tan(x) term.

@edmundwoolliams1240 9 ай бұрын

So would you just insert the series expansion of tan as well, then solve the slew of multiplied infinite sums to get some kind of recurrence relation do you reckon? @@maths_505