a logarrific differential equation

[@ELKCreativist]

Quick notice: y' can't be zero in the original DE (y''/y' - y'/y = lny), thus case 1 should've been automatically rejected.

[@drpeyam]

Unless you accept the convention here that 0/0 = 0

[@TheAleBecker]

heresy ​@@drpeyam

[@JohnVKaravitis]

@@drpeyam 0.0 is undefined, so.....

[@drpeyam]

In this example I would say it’s valid, at least for the multiplied version

[@marigold2257]

@@drpeyamadding 1/1 to both sides we have 1/1+0/0=0+1/1=1 1/1+0/0=((1*0)+(1*0)/1*0=0/0=1 0/0=0 0=1 In no world does 0/0=0 please try to check that things are true before spreading misinformation

[@mauricebre4969]

Or you can write ln(y')'-ln(y)'=ln(y) iff ln(y'/y)'=ln(y) than substitute u=ln(y) than you get u''=u*u'=1/2(u^2)' and this is really easy to solve. Ps. Really nice ODE and really nice video.

[@r.maelstrom4810]

How in the hell u'' = u*u'? u'' = (y''y-(y')^2)/y^2 and uu'=ln(y)y'/y there's no way they are equal.

[@mauricebre4969]

@@r.maelstrom4810 With u’=ln(y)’=y’/y=y’/e^u => ln(y’)=ln(u’)+u Put this in the ODE: (ln(u’)+u)’-u’=ln(u’)’=ln(y)=u iff u’’/u’=u iff u’’=u*u’ About your technique: u’’=u*u’ iff (y’’y-(y’)^2)/y=ln(y)y’ iff(Solve the bracket and divide both sides by y’) y’’/y’-y’/y=ln(y) And you see the original ODE

[@alejrandom6592]

I kept having this deja vu that i had tried to solve this before, and in fact Michael Penn has a video on this ODE, using a different method. It's called "A nice suggested differential equation" for those interested.

[@drpeyam]

Interesting

[@AdrianRif]

Hi Dr Peyam, I have a much more elegant solution. Rearrange equation so that y’’ - y’ = y ln y Integrate both sides. On RHS, sub u = ln y, so that y = exp u. Then y’ - y = exp u ( u - 1) + C = y ln y - y + C ; constant C so rearranging we get y’ = y ln y + C Integrate both sides again and use same substitution on right hand side y = y ln y - y + Ct + B; constant B rearrange 2y - y lny = Ct + B. Nice one eh?

[@AdrianRif]

Sorry I didn’t notice the y’ in the denominator!

[@omograbi]

You've missed a "dw" at the end of the last integral <a href="#" class="seekto" data-time="633">10:33</a>

[@andrewparker8636]

You can just use the substitution z=lny to simplify the original equation.

[@drpeyam]

Not as obvious!

[@alejrandom6592]

We can start by noticing that y'/y = d/dt (ln(y)) and y"/y' = d/dt (ln(y')) We have d/dt (ln(y')-lny) = lny d/dt (ln(y'/y)) = lny If we let u=lny Then y'/y=u' Substituting we get d/dt [ln(u')] = u u"/u' = u u" = uu' du/dt = (1/2)u²+c And from there you solve it as the video In the end a simpler way to relabel constants is y = exp(2A*tan(At+B))

[@drpeyam]

I like this!!

[@thelittlesillystar]

i was thinking of differential equations this morning so seeing this is pretty awesome! crazy result as well

[@drpeyam]

Glad you liked it!

[@akirakato1293]

btw the substitution u(y)=y' can you only do if there are only y terms in the differential equation? because otherwise how do you know that y' can be expressed with only y?

[@holyshit922]

Suppose you have fellow and you want to give him ode to solve I can give yo two recipes 1. Choose p(x), q(x) and form Bernoulli equation of degree 2 Choose y1(x) and substitute to get Riccati equation from Bernomulli equation Use subsitiution to get second order liniear from Riccati equation 2. Pick a, b,c , f(t) and write second order linear equation Use t = g(x) substitution to get more complicated equation

[@anandarunakumar6819]

I would have started with U(y) = y'/y, then proceed to get similar result. Recently did almost a similar integral of this kind!

[@cheeseparis1]

Left handed math teacher adds even more suspense

[@alejrandom6592]

<a href="#" class="seekto" data-time="838">13:58</a> shouldn't tan be in the exponent?

[@drpeyam]

True!!

[@alejrandom6592]

I thought I was going crazy for a moment

[@allahkaramshafie4018]

That's brilliant

[@allahkaramshafie4018]

Dear Pyam, how can I connect you ? Thanks. Arash

[@holyshit922]

This equation can be reduced to the first order Standard approach is to use substitution y' = u(y) Maybe example for change of independent variable next time Method which i mentioned usually not appear in US ODE class

[@drpeyam]

Yes I never heard of this method actually!

[@holyshit922]

@@drpeyam For exaple you have equation y''(t)+p(t)y'(t) + q(t)y(t) = 0 and you put x = g(t) to get equation y''(x) + P(x)y'(x) + Q(x)y(x) = 0 Another example Suppose that we want to derive ODE for Chebyshev polynomials We know that T_{n}(x) = cos(n*arccos(x)) Let y(t) = cos(nt) If we differentiate it twice we will get y''(t)+n^2y(t) = 0 If we put x = cos(t) we will get equation (1 - x^2)y''(x) - xy'(x) + n^2y(x) = 0 Now if we want to get coefficients of Chebyshev polynomial we can solve for polynomial solution satisfying condition y(1) = 1

[@leonardobarrera2816]

How did you meet BPRP?

[@alejrandom6592]

The intonet

[@leonardobarrera2816]

@@alejrandom6592 eeee?

[@giuseppemalaguti435]

Risulta (lny')'-(lny)'=lny...posto lny=u..y=e^u,y'=e^u*u'..lny'=u+lnu'..perciò risulta (lnu')'=u,cioè u"/u'=u..u"=u*u'..(u')'=(1/2u^2)'..cioè u'=u^2/2+c..poi e' semplice...

[@pasha5932]

Lovely procedure 😁

[@drpeyam]

Thank you! 😊

[@wagsman9999]

Nice

[@alipourzand6499]

Is the the zombi outbreak ODE that you talked about?☺

[@drpeyam]

Hahaha yes sure 😂

[@Galileosays]

When y=C then y'=y''=0 and the original equation fails: 0/0-0/C=ln(C) ??? Seems a division by zero problem.

[@ostorjulien2562]

Or let z=ln y, then z''=z x z' and we are (almost) done

[@r.maelstrom4810]

How in the hell z'' = z*z'? z'' = (y''y-(y')^2)/y^2 and zz'=ln(y)y'/y there's no way they are equal.

[@ostorjulien2562]

@@r.maelstrom4810 the inital equation can be writen (ln(y') - ln(y))' = ln(y). If z = ln(y), then y = e^z and y' = z' e^z so ln(y') = ln(z') + z. Replacing in the rewriten equation leads to (ln(z') + z - z)' = z. Simplifying : z''/z = z'. Finally, I got two solutions: y(x) = exp(a tan(2ax + b)) and y(x) = exp(a tanh(-2ax + b), a and b are constants.

[@actions-speak]

Just what I tried! It's a very satisfying way to solve the problem.

[@samueldeandrade8535]

​@@ostorjulien2562 how do you know you can take ln(y') ??? I mean, to do that you need to know y'(x) > 0, for all x How do you know that?

[@actions-speak]

@@samueldeandrade8535 I don't think that's necessary. Let z = ln(y). Then y = e^z, y' = z' e^z, y'' = z'' e^z + (z')^2 e^z. Substituting into the given equation, (z'' + (z')^2)/z' - z' = z and z'' = z' z.

[@tomaszkochaniec9421]

Mayby faster is y"/y'=(ln(y'))' and y'/y=(ln(y))'

[@drpeyam]

I was thinking that too but it didn’t lead me anywhere sadly

[@tomaszkochaniec9421]

@@drpeyam what if we integrate on both sides?

[@mnek742]

@@tomaszkochaniec9421 you'd have to integrate ln(y) with respect to t then, can't really go too far with that

[@tomaszkochaniec9421]

@@mnek742 try to Taylor series log(y) and all clear.

[@mnek742]

​​@@tomaszkochaniec9421 I think you may be confusing integrating with respect to y with integrating with respect to t. The integral of ln(y) wrt y is not hard, in fact it's y*ln(y)-y+C, no Taylor series needed. Integrating ln(y) wrt t is not really possible even with a Taylor expansion. Here it helps to have a real-world analogy and understand what it means to integrate. Suppose y was the population of some town, and t was time in months, and you wished to integrate ln(y). Roughly speaking that would loosely entail tabulating and adding values of ln(y) month by month, just straight addition. But we have no direct knowledge (yet) of how the town's population is growing so we don't know what we're adding. So the integration can't really be done, you would just write (integral) ln(y) dt and it wouldn't help solve the DE

[@marccox8977]

Dr. P Use the Chen Lu!! 😁👍

[@HistorymanDublin]

Is there any real world case with that equation?

[@CKSLAFE]

this

[@HistorymanDublin]

A real world situation where this equation is useful? Does the equation have any application in physics etc.?

[@thohoduc5987]

I am sorry Dr. Peyam, but I think there is a mistake in your solution. The step y''(t) = (dU/dy)*y'(t) is not correct for an assumed general function y(t). The function y = t^2 is an example.

[@drpeyam]

It is correct actually, by the chain rule

[@thohoduc5987]

​@@drpeyam Thank you for your clarification. I got it.

[@jamesromano3288]

He got so skinny.....is he ok "?"?"?

[@tszulpinedo757]

Hello, Dr. Peyam n.n/

[@drpeyam]

Hiiiii!!!

[@subscriberunavailable]

You look different these days it's like you have seen a ghost

[@BikramSingh-c7o]

Easy pessy

[@JohnVKaravitis]

THIS is why I hate math.