(Differential Equations) Determining Linearity of a Function

electronicaneer
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This is more of a general question than a specific homework question, because it popped up in more than 1 problem. If you have 'x' has the independent variable and 'y' as the dependent variable, you can determine the linearity in 'y' by seeing if any of the derivatives (dy/dx) are being raised to a power or not.

How would you go about determining the linearity in 'x' then though? Would you rearrange it so all of the derivatives reflect (dx/dy) and then re-evaulaute?
 
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I suppose you could but generally, unless you're talking about first order equations, it would be hard and pointless to do. For example, think of a DE which gives a particles trajectory x as a function of time t. In this scenario, it would be perfectly fine to have two different times correspond to one position. However upon flipping it, and trying to make t as a function of x, you would have one position corresponding to two different times (this tells you that t really is not the dependent variable) . The example shows that although sometimes you can (in simple cases), it might produce mathematically inconsistencies or just be impossible to do analytically

Also, as a side note, what you're trying to determine in differential eq is the linearity of the differential operator on the solution (y) and not the linearity in y itself
 
There is no reason to define "linear in x" for a differential equation because it has no effect on how the equation might be solved. A "linear differential equation" is one that is "linear in y". By the way, you say "you can determine the linearity in 'y' by seeing if any of the derivatives (dy/dx) are being raised to a power or not." I presume you intended to include y itself in "any of the derivatives". The equation dy/dx= y^2 is non-linear even though all of the derivatives are not to a power. And of course, you should include other, non-polynomial, functions. dy/dx= sin(y) and d^2y/dx^2+ e^{dy/dx}+ y= 0 are non-linear differential equations.
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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