My Math Forum Why this diff. equation does not have a degree?

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 July 3rd, 2016, 07:54 AM #1 Newbie   Joined: Oct 2014 From: 192.168.0.1 Posts: 13 Thanks: 2 Why this diff. equation does not have a degree? In a book, it was stated that the following equation is of second order but no degree since the equation cannot be rationalized. $\displaystyle t\left ( \frac{\mathrm{d^2} y}{\mathrm{d} t^2} \right )+t^{2}\frac{\mathrm{d} y}{\mathrm{d} t}-\left ( cost \right )\sqrt{y}=2t^{2}-3t+4$ Why cannot we rearrange the equation as follows: $\displaystyle t\left ( \frac{\mathrm{d^2} y}{\mathrm{d} t^2} \right )+t^{2}\frac{\mathrm{d} y}{\mathrm{d} t}-2t^{2}+3t-4=\left ( cost \right )\sqrt{y}$ Then, square both the sides to get the following: $\displaystyle \left ( t\left ( \frac{\mathrm{d^2} y}{\mathrm{d} t^2} \right )+t^{2}\frac{\mathrm{d} y}{\mathrm{d} t}-2t^{2}+3t-4 \right )^{2}=\left ( cos^{2}t \right )y$ Can't we say that the above equation is of second order and second degree?
 July 4th, 2016, 09:07 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 The equation, as it stands, is not a polynomial in y or its derivatives because of the $\displaystyle \sqrt{y}$. Squaring the equation eliminates the square root but then it is not the same equation, just as x= 1 is not the same equation as $\displaystyle x^2= 1$!

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