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In my differential equations book, we are trying to prove the linear dependence of a set of $3$ functions. We start with

$$c_1x^2+c_2x^3+\dfrac {c_3}{x}=0$$

Then the book says to differentiate twice, and get the following two equations

$$2c_1x+3c_2x^2-\dfrac{c_3}{x^2}=0$$

$$2c_1+6c_2x+\dfrac{c_3}{2x^3}=0$$

We set $x=1$ and go on to solver for $c_1, c_2, c_3$ and show that they all must be $0$.

However, why are we allowed to differentiate equations containing only one variable? If I have the equation $2x=5$ and differentiate both sides, I get $2=0$. The only way out I can see is if we temporarily consider $c_1, c_2, c_3$ as variables, but if that were the case then why aren't we seeing any $\dfrac {dc_1}{dx}$?

Ovi
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  • functions or equations ? – Claude Leibovici Mar 13 '16 at 04:59
  • @ClaudeLeibovici We are given $3$ solutions to a third order DE, namely $y_1=x^2$, $y_2=x^3$, and $y_3=1/x$ and we are asked to prove that these solutions are linearly independent – Ovi Mar 13 '16 at 05:08
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    $2x=5$ is an equation with the solution $x=2.5$. It isn't a function though, so its derivative isn't well defined. Consider $f(x)=2x-5$. If you want to find when $f(x)=0$, you get back your old equation. Now we have a function though, so we can take its derivative (namely $f'(x)=2$

    The issue is that in terms of functions you have $2x$ and $5$ being equal at one point, but for two functions to be equal they must agree at all points.

     – Mark Schultz-Wu Mar 13 '16 at 05:10
  • @Mark But why is the book allowed to differentiate that equation? It is also just an equation with only one variable, $x$, with $c_1, c_2, c_3$ as constants – Ovi Mar 13 '16 at 05:16
  • @Ovi: Because it is an equation of functions (i.e., the equation is true for all $x$), not just an equation that is supposed to hold for a single value of $x$. See the answers at the question I linked above for more discussion. – Eric Wofsey Mar 13 '16 at 05:19

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