towards the notion of an exact second order differential equation

[@coreyyanofsky]

13:36 seems like there's a mistake here because (x²y)' = 2xy + x²y' and not what's written on the board

[@therealbrewer]

He needs to swap the 2x and the x^2 in the top equation.

[@goodplacetostop2973]

16:42

[@Pedritox0953]

It would be a plus to expand more about the foundations of this problem, Great video! Peace out

[@GreenMeansGOF]

Please show us the general equation and method.🤩

[@koengroot3285]

The general 2nd-order equation appears to be transformed to a Riccati equation in y'?

[@holyshit922]

Coefficients depend on y also so I dont think so
but you can transform second order linear to Riccati
y''(x)+p(x)y'(x)+q(x)y(x) = 0
y''(x) = -p(x)y'(x) - q(x)y(x)
y''(x)/y(x) = -p(x)(y'(x)/y(x)) - q(x)
y''(x)/y(x) - (y'(x)/y(x))^2 = -(y'(x)/y(x))^2 -p(x)(y'(x)/y(x)) - q(x)
y''(x)/y(x) - y'(x)y'(x)/y(x)^2 = -(y'(x)/y(x))^2 -p(x)(y'(x)/y(x)) - q(x)
(y''(x)y(x) - y'(x)y'(x))/y(x)^2 = -(y'(x)/y(x))^2 -p(x)(y'(x)/y(x)) - q(x)
d/dx(y'(x)/y(x)) = -(y'(x)/y(x))^2 -p(x)(y'(x)/y(x)) - q(x)
Let y'(x)/y(x) = z(x)
You have
z'(x) = -z^2(x)-p(x)z(x)-q(x)
y'(x) = y(x)z(x)
So second order linear can be transformed to Riccati

[@holyshit922]

This transformation usually is not taught in differential equations class because it is rarely useful
You still need particular solution but sometimes particular solution can be easier guessed after reduction to Riccati
For second order linear it is possible

[@koengroot3285]

@@holyshit922 Yes, good point. I think the Riccati form of the transformed equation is not a coincidence. The approach to the exact differential equation of first order is to achieve a reduction of (differential) order (to zeroth order, so that psi(x,y) = cst). This same objective leads us to transform a second-order ODE to a Riccati equation.

[@holyshit922]

@@koengroot3285
p(x,y)y''+q(x,y)(y')^2+r(x,y)(y)'+s(x,y) = 0
is for you Riccati equation in y' ?
for me it is p(x)y''+q(x)(y')^2+r(x)(y)'+s(x) = 0
Do you see the difference

[@koengroot3285]

​@@holyshit922Yes, the equation with "coefficients" dependent on both x and y is what I referred to as a Riccati equation in y'; indeed, I realized it's quite a bit more general than the Riccati equation that results from reducing the differential order of a second-order ODE

[@writerightmathnation9481]

It would be nice to have some references, Michael. I'll look it up, but I taught differential equations and I don't recall seeing this in any texts; I don't do research at the cutting edge of DiffEq, but knowing where you found it would be of interest to some viewers (me, for instance...).

[@TheLiuzp]

Is there any physical meaning for this 2nd order exact DE? Or it's just for the sake of generalizing exact DE's?
I mean, a 1st order exact DE can be related to the physics concept that -grad(V) = F, where V is a potential function and F a force field, and what about a 2nd order exact DE?

[@richardchapman1592]

I fancy that the tangent at x,y can be refined to make the limiting processes of differentiation converge more swiftly at extremes of curvature. May be able to smooth discontinuities a little to make them treatable as pseudo continuous.

[@Czeckie]

exact differential equation means that we the given 1-form is exact. Is there a differential geometry/tensor calculus way of stating this 2nd order exact de?

[@Christopher-e7o]

X,2x+5=8

[@imtiyazansary]

At 13:36 you did a silly mistake

[@holyshit922]

If you want to practise then from earlier Michael Penn's video
ruclips.net/video/oTDBz9G5yEs/видео.html

[@charleyhoward4594]

1 st time I've every seen Micheal unshaved ... not becoming

[@Alan-zf2tt]

second 🙂