Linearity in Differential Equations

TranscendArcu
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Homework Statement



Is the following differential equation linear:

yy' + 2 = 0

The Attempt at a Solution



I have the definition of linear as being a_0 (t) y^{(n)} + a_1(t) y^{n-1} + a_2 (t) y^{n-2} ... = 0. Now, presumably y is a function of t. Thus, I could define y = a_0 (t) and let n=1. Thus I would satisfy my definition of linearity in differential equations. Thus, the differential equation is linear.

Is it not so?
 
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No, a_0(t) has to be a known function of t, not y or anything involving y. This equation is non-linear because you have a product of things "involving" the dependent variable y.
 
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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