a logarrific differential equation 

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

heresy ​@@drpeyam

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

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

@@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

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.

@@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

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

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

@@drpeyam what if we integrate on both sides?

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

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

​​@@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

this

Thank you! 😊

It is correct actually, by the chain rule

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

Glad you liked it!

Hahaha yes sure 😂

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.

@@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.

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

​@@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?

@@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.

Hiiiii!!!

Yes I never heard of this method actually!

@@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