My try :
I have taken $y_1,y_2$ and tried to get a recursive relation between them but couldn't find any pattern. Please help.
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you have not specified what $y_n$ is – RE60K Aug 23 '14 at 07:14
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1hint : y^(1/m)=a , so , a+1/a =2x, you can find a (y ^(1/m)) from that by multiplying a , and solve quadratic equation – Khosrotash Aug 23 '14 at 07:22
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I think @Aditya $y_n$ is $n$th derivative but the textbook doesn't explicitly mention anything. – AgentS Aug 23 '14 at 07:29
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@daryakhosrotash that looks new ! I'll try solving y^(1/m) using quadratic formula :) thanks ! – AgentS Aug 23 '14 at 07:30
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Applying derivative $$\frac{y^{\frac1m-1}}my_1-\frac{y^{-\frac1m-1}}my_1=2 \implies\left(y^{\frac1m}-y^{-\frac1m}\right)=\frac{2my}{y_1}$$
Now, $\displaystyle\left(y^{\frac1m}-y^{-\frac1m}\right)^2=\left(y^{\frac1m}+y^{-\frac1m}\right)^2-4$
$\displaystyle\iff \left(\frac{2my}{y_1} \right)^2=\left(2x\right)^2-4$
$$\iff m^2y^2=(x^2-1)y_1^2$$
Applying derivative $\displaystyle m^2(2yy_1)=2xy_1^2+(x^2-1)2y_1y_2$
Cancelling $2y_1$(assuming $\ne0$)
$$\iff m^2y=xy_1+(x^2-1)y_2$$
Now apply General Leibniz rule
lab bhattacharjee
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Thanks a lot ! I have understood till the last line. But not clear on which two functions to pick for leibniz rule. Should I pick the product $xy_1$ ? – AgentS Aug 23 '14 at 07:56
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@rsadhvika, See http://wdjoyner.com/teach/calc1-sage/html/node106.html or http://en.wikipedia.org/wiki/Product_rule#Higher_derivatives – lab bhattacharjee Aug 23 '14 at 08:47
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I had used leibniz rule several times before for taking nth derivative of product of funcitons. I'm not able to apply it corretly here, I'm still struggling... – AgentS Aug 23 '14 at 08:51
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1@rsadhvika, $$\frac{d^n{y_2(x^2-1)}}{dx^n}=(x^2-1)\frac{d^n(y_n)}{dx^n}+\binom n1\frac{d(x^2-1)}{dx^1}\frac{d^{n-1}(y_n)}{dx^{n-1}}+\binom n2\frac{d^2(x^2-1)}{dx^2}\frac{d^{n-2}(y_n)}{dx^{n-2}}+0$$ as $$\frac{d^m(x^2-1)}{dx^m}=0$$ for $m\ge3$ – lab bhattacharjee Aug 23 '14 at 08:55
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