Gwendolyn Willett

Answered

2022-01-15

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

Answer & Explanation

Jeffery Autrey

Expert

2022-01-16Added 35 answers

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 $y}^{\left(n\right)}=Ak(k-1)\dots (k-n){x}^{k-n$ . This gives the formula ${A}^{2}k(k-1)\dots (k-n){x}^{2k-n-1}=1$

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

$A={(k(k-1)\dots (k-n))}^{-\frac{1}{2}}$

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.

For n an odd number, try

which can only be true if

Furthermore one has to have that

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

Dabanka4v

Expert

2022-01-17Added 36 answers

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

alenahelenash

Expert

2022-01-24Added 366 answers

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 $x=f(y),\text{so}\text{}y={f}^{-1}(x)$ Then $dx={f}^{\prime}(y)dy$ ,or $\frac{dy}{dx}=\frac{1}{{f}^{\prime}(y)}$ .So $\frac{{d}^{2}y}{d{x}^{2}}=\frac{d}{dy}(\frac{1}{{f}^{\prime}(y)})\frac{dy}{dx}=-\frac{{f}^{\u2033}(y)}{({f}^{\prime}(y){)}^{2}}\frac{dy}{dx}=-\frac{{f}^{\u2033}(y)}{({f}^{\prime}(y){)}^{3}}=-\frac{{g}^{\prime}(y)}{(g(y){)}^{3}}$ where $g(y)={f}^{\prime}(y)$ .Solving $-\frac{{g}^{\prime}(y)}{(g(y){)}^{3}}=\frac{1}{y}$ , we have$g(y)={f}^{\prime}(y)=\frac{1}{\sqrt{\mathrm{log}({y}^{2})+c}}$ Now we need to solve for f(y), and invert the function to get $y={f}^{-1}(x)$

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