Gwendolyn Willett

2022-01-15

Solutions to the equation ${y}^{n}y=1$ for even n

Jeffery Autrey

Expert

A long time ago I was curious about the closed-form solutions to the equation:
$\frac{{d}^{n}y}{{dx}^{n}}y=1$
For n an odd number, try $y=A{x}^{k}$. Then ${y}^{\left(n\right)}=Ak\left(k-1\right)\dots \left(k-n\right){x}^{k-n}$. This gives the formula ${A}^{2}k\left(k-1\right)\dots \left(k-n\right){x}^{2k-n-1}=1$
which can only be true if $k=\frac{n+1}{2}$ and n is odd (k cannot be an integer for the formula to work, check this youself.)
Furthermore one has to have that
$A={\left(k\left(k-1\right)\dots \left(k-n\right)\right)}^{-\frac{1}{2}}$
which is real if n is odd.
Thus there are closed-form solutions to my problem for n odd, and my question is if anyone can find a closed-form solution for $n=2$ or in general if n is even.

Dabanka4v

Expert

Let $y=y\left(x\right)$ and write s

alenahelenash

Expert

Well i tried for $\frac{{d}^{2}y}{d{x}^{2}}y=1$, and i have partial solution. I will use this reply to trace the work since it supports LaTeX, I will try to complete it later (or someone else can take over).Suppose Then $dx={f}^{\prime }\left(y\right)dy$,or $\frac{dy}{dx}=\frac{1}{{f}^{\prime }\left(y\right)}$.So $\frac{{d}^{2}y}{d{x}^{2}}=\frac{d}{dy}\left(\frac{1}{{f}^{\prime }\left(y\right)}\right)\frac{dy}{dx}=-\frac{{f}^{″}\left(y\right)}{\left({f}^{\prime }\left(y\right){\right)}^{2}}\frac{dy}{dx}=-\frac{{f}^{″}\left(y\right)}{\left({f}^{\prime }\left(y\right){\right)}^{3}}=-\frac{{g}^{\prime }\left(y\right)}{\left(g\left(y\right){\right)}^{3}}$ where $g\left(y\right)={f}^{\prime }\left(y\right)$.Solving $-\frac{{g}^{\prime }\left(y\right)}{\left(g\left(y\right){\right)}^{3}}=\frac{1}{y}$, we have$g\left(y\right)={f}^{\prime }\left(y\right)=\frac{1}{\sqrt{\mathrm{log}\left({y}^{2}\right)+c}}$Now we need to solve for f(y), and invert the function to get $y={f}^{-1}\left(x\right)$