I am given $$ y = \alpha + \frac{1}{\alpha} $$ and asked to prove that $$ \alpha^2 y'' + \alpha y-2 = 0 $$ but when I use $\alpha$ as a variable I get $y-2 = 0$. Am I missing something? No other explanation is given regarding $\alpha$ or $y$.
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It is wrong. $y$ is a function of $\alpha$. We have $y'=1-\frac 1 {\alpha^{2}}$ and $y''=\frac 2 {\alpha^{3}}$. Hence $\alpha^{2}y''+\alpha y-2=\frac 2 {\alpha}+\alpha^{2}+1-2$ which is not $0$.
Kavi Rama Murthy
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Sorry, the diferential equation is in fact $$ \alpha^2 y'' + \alpha y' - 2 = 0 $$ and I thought it was probably a special kind of equation. – Jesús A. Piñera Mar 13 '21 at 05:47
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@JesúsPiñera Even then you don't get $0$. – Kavi Rama Murthy Mar 13 '21 at 05:53