1

I'm trepidatious to accept an answer here because there's a couple points the author hasn't cleared up

Why doesn't this differential technique work?

$y''/y$ certainly exists if the solution exists wherever $y \neq 0$, and it is integrable.

Secondly, $y''/y = t$ which is an invertible function. In the worst case scenario, suppose we define a new function, the inverse Airy function's derivative over a compact set $Ai'^{-1}(t).$

We can represent $t$ then as $Ai'^{-1}(Ai'(t)) = t$ over varying compact intervals corresponding to the appropriate branch of its inverse, so clearly an antiderivative does exist in terms of $y',$ though it is then still an open-question as to how we derive it.

What are the flaws in our reasonings here?

StackQuest
  • 579
  • You want to integrate $y''/y$ to get a solution of $y''=ty$ ? – Kurt G. Mar 05 '22 at 09:39
  • It would be interesting to know if there was a concise way to integrate $y'' / y$ to obtain a differential relationship in terms of $y$ and $y'.$ – StackQuest Mar 06 '22 at 18:28

1 Answers1

1

In a nice answer to the post you are referring to it was demonstrated that $y''/y$ is not the first derivative of a function $f(y,y')\,.$ That's the reason we can't separate variables when we solve the ODE $y''=y\,g(x)\,.$

This is maybe not the end of the world. Apparently, $$ \frac{y''}{y}-\frac{y'^2}{y^2} $$ is the second derivative of the function $f(y)=\log y\,.$ More generally, for arbitrary twice differentialbe $f\,,$ $$ \frac{d}{dx}f(y)=f'(y)\,y'\,,\quad\frac{d^2}{dx^2}f(y)=f'(y)\,y''+f''(y)\,y'^2\,. $$ When $f$ is invertible this means that the ODE \begin{align}\tag{1} f'(y)\,y'&=g(x)\,,\\[3mm] \end{align} has the solution \begin{align} \textstyle y=f^{-1}\Big(\int_0^x g(z)\,dz\Big)+\Big(y(0)-f^{-1}(0)\Big)\,. \end{align} For the ODE \begin{align} f'(y)\,y''+f''(y)\,y'^2&=g(x)\tag{2} \end{align} we get first $$ f'(y)\,y'=\textstyle\int g(x)\,dx=:h(x) $$ which is an ODE of the form (1) and has a known solution.

I believe the old Book Differntialgeleichungen by E. Kamke (not sure if it was ever translated) is a good reference for such "non ordinary" ODEs.

Kurt G.
  • 14,198
  • Okay, so you've introduced this notion of "separable," what does that mean in an ODE? Because if you construct arbitrary special functions, you could always rearrange an equation to isolate $y$ or $y'.$ $y'' = g(x)y$ looks separable to me, but you're saying the integral of this expression is some type of non-separable combination of $y'$ and some $\int g(x)$? – StackQuest Mar 07 '22 at 00:44
  • Also, $\int y'' / y = \frac{1}{2} t^2$ (or $\frac{1}{2} x^2$ in your case). If you define a special function $Ai'^{-1}(t),$ you can represent this integral as $Ai'^{-1}(Ai'(t))^2,$ so it appears to me there does exist an explicit expression in terms of $y'.$ – StackQuest Mar 07 '22 at 00:49
  • Okay, separable doesn't seem like a huge deal. So, what if I don't care whether the integral of $y''/y$ is separable in terms of $y'$ and $y,$ but simply exists in some closed form in terms of $y'$ and $y$ and $t$ (or $x$)? It could be the least separable thing you can imagine like $\int \Gamma(y^{y'} - log_{y}(y'))$ or etc, but as long as it exists at all, I'm completely fine with that, I don't see why we make a big deal out of being separable, so please don't restrict yourself to that. – StackQuest Mar 07 '22 at 04:27
  • To your first comment : $y''=g(x)y$ is separable but -as the nice answer in your related question demonstrates- we cannot use that version of separability to find the solution of the ODE. To your second comment: whatever you do to make $t^2$ look different won't help, because again: see answer to first comment. To your third comment: how do you want to solve $\int\Gamma(y^{y'}-\log_y(y'))$ for $y$ ? – Kurt G. Mar 07 '22 at 10:06
  • So what "version" of separable are you suggesting? I don't mind if we "can't" use that version, why would I? I showed $t^2$ can be written in terms of the second derivative, so my question is half-answered. I don't know how to integrate the gamma function, but the result is some other special function in terms of $y'$ and $y$ that must admit some functional inverse relationship that lets one solve for $y$ locally. – StackQuest Mar 08 '22 at 03:06
  • I suggest the "version" which we name after Joseph Fourier (1768-1830) since more than two hundret years. – Kurt G. Mar 08 '22 at 07:18