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I've been playing around with differential equations. I can easily solve the differential equation $$y'\cdot y=1$$ for $y:\mathbb R\mapsto\mathbb R$, and I can also solve $$y''\cdot y=1$$ using substitution into the previous example. However, I cannot figure out this differential equation: $$y''\cdot y'\cdot y=1$$ Does anybody have any ideas about how to solve this?

So far, the only technique that I know that seems valid for this differential equation is the use of Taylor Series, but that gets too messy for me to get anything useful out of it.

Franklin Pezzuti Dyer
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    One approach would be to treat $y$ as the independent variable and substitute $\frac{dy}{dx}=v(y)$ giving $\frac{d^2 y}{dx^2}=v(y)\frac{dv}{dy}$ by the chain rule. You will obtain a separable ODE: $$\frac{dv(y)}{dy}=\frac{1}{y\cdot v(y)^2}$$ – projectilemotion Nov 02 '17 at 23:13
  • I'd like to see how you solve $y''y = 1$ – Dylan Nov 05 '17 at 19:53
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    @Dylan You could probably do it by making the same substitution that projectilemotion made. – Franklin Pezzuti Dyer Nov 07 '17 at 00:21

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Considering the differential equation $$\frac{d^2y}{dx^2}\,\frac{dy}{dx}\,y=1$$ let us use $$\frac{dy}{dx}=\frac1 {\frac{dx}{dy}}\qquad \text{and}\qquad \frac{d^2y}{dx^2}=-\frac{\frac{d^2x}{dy^2}}{\left(\frac{dx}{dy}\right)^3 }$$ (see here),we then arrive to $$x'' y=-(x')^4$$ Now, using $u=x'$, this makes $$u' y=-u^4\implies \frac{u'}{u ^4}=-\frac 1y\implies \frac{1}{3 u^3}=\log(y)+c_1$$ which has three solutions in $u$.

Considering one of them $$u=\frac{1}{\sqrt[3]{3} \sqrt[3]{\log (y)+c_1}}$$ $$x=-\frac{e^{-c_1} }{\sqrt[3]{3}}\Gamma \left(\frac{2}{3},-(c_1+\log (y))\right)+c_2$$ from which $y$ cannot be extracted.

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Claude Leibovici
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$$y''=\frac{dy'}{dy}\frac{dy}{dx},$$ $$y''y'y=y'^2\frac{dy'}{dy}y=\frac13\frac{dy'^3}{d\ln y}.$$ You should be able to finish the rest.

Hans
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    But not to an explicit solution. – Robert Israel Nov 02 '17 at 23:19
  • @RobertIsrael: If by "explicit" you mean an algebraic function, then that is true. However, would you not an integral to be a legitimate solution? Had it been a, say, Gamma or Zeta function which has integral representation in some domain, would you consider it "explicit" enough? They are all integrals, expect Gamma and Zeta have names given to them. – Hans Nov 02 '17 at 23:29
  • However, this integral gives a solution of the form $F(y) = x$, which is still implicit rather than explicit. – Robert Israel Nov 03 '17 at 00:25
  • @RobertIsrael: Yes, but why is this objectionable? Any ODE $\frac{dy}{dx} = f(x)$ which gives an explicit solution can be turned into one $\frac{dx}{dy}=f(y)$ that gives an implicit solution. Besides, a function is just a mapping between coordinates. Mapping from one coordinate to another or vice versa is just an arbitrary choice. What essential difference does it make? We have implicit function theorem at our disposal anyway if we desire rigor. – Hans Nov 03 '17 at 07:21