If $y^{1/m} + y^{-1/m} = 2x$ then prove that $(x^2 - 1)y_{n+2} + (2n + 1)x y_{n+1} + (n^2 - m^2)y_n = 0$?
Where $y_n$ denotes the $n^th$ derivative of $y$. This is a question of successive differentiation and has to be proved using Principle of mathematical induction. But I seem to make no considerable progress while attempting to solve this using successive differentiation.
The question I wrote I same as mentioned in my textbook. I'm wondering without mentioning anything about the variables n, m how can it be solved. Is there any approach. Is there anything that one can assume to proceed further.
This is a problem from single variable calculus text. I'm a beginner to single variable calculus. I have tried it using successive differentiation rule of leibnitz but unable to reach my answer.
This approach might work:
$\frac d{dx} \left(y^{\frac 1m}\right )+\frac d{dx} \left(y^{-\frac 1m}\right )=\frac d{dx} \left(2x\right )$
$\frac 1m \left(y^{\frac 1m -1}\right )\frac {dy}{dx}-\frac 1m \left(y^{-\frac 1m -1}\right )\frac {dy}{dx}=2$
$\left(y^{\frac 1m}-y^{-\frac 1m}\right )\frac {dy}{dx}=2my$
$\frac d{dx}\left(y^{\frac 1m}-y^{-\frac 1m}\right )\frac {dy}{dx}+\left(y^{\frac 1m}-y^{-\frac 1m}\right )\frac {d^2y}{dx^2}=2m\frac {dy}{dx}$
$\left(\frac 1m \left(y^{\frac 1m -1}\right )+\frac 1m \left(y^{-\frac 1m -1}\right )\right )\frac {dy}{dx}+\left(y^{\frac 1m}-y^{-\frac 1m}\right )\frac {d^2y}{dx^2}=2m\frac {dy}{dx}$
$\left(y^{\frac 1m} + y^{-\frac 1m} \right )\frac {dy}{dx}+my\left(y^{\frac 1m}-y^{-\frac 1m}\right )\frac {d^2y}{dx^2}=2m^2y\frac {dy}{dx}$
$\left(y^{\frac 1m} + y^{-\frac 1m} \right )\left(\frac {dy}{dx}\right)^2+my\left(y^{\frac 1m}-y^{-\frac 1m}\right )\frac {d^2y}{dx^2}\frac {dy}{dx}=2m^2y \left(\frac {dy}{dx}\right)^2$
$2x\left(\frac {dy}{dx}\right)^2+2m^2y^2\frac {d^2y}{dx^2}=2m^2y \left(\frac {dy}{dx}\right)^2$
Use these to prove thet statement true for $n=0$ then proceed by induction.