Solutions to the equation yny=1 for even n

Gwendolyn Willett

Gwendolyn Willett

Answered

2022-01-15

Solutions to the equation yny=1 for even n

Answer & Explanation

Jeffery Autrey

Jeffery Autrey

Expert

2022-01-16Added 35 answers

A long time ago I was curious about the closed-form solutions to the equation:
dnydxny=1
For n an odd number, try y=Axk. Then y(n)=Ak(k1)(kn)xkn. This gives the formula A2k(k1)(kn)x2kn1=1
which can only be true if k=n+12 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(k1)(kn))12
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

Dabanka4v

Expert

2022-01-17Added 36 answers

Let y=y(x) and write s
alenahelenash

alenahelenash

Expert

2022-01-24Added 366 answers

Well i tried for d2ydx2y=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),so y=f1(x)Then dx=f(y)dy,or dydx=1f(y).So d2ydx2=ddy(1f(y))dydx=f(y)(f(y))2dydx=f(y)(f(y))3=g(y)(g(y))3 where g(y)=f(y).Solving g(y)(g(y))3=1y, we haveg(y)=f(y)=1log(y2)+cNow we need to solve for f(y), and invert the function to get y=f1(x)

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