Differential equations true/false question

freshman2013
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Homework Statement


If y1 and y2 are solutions to y"-y = 0 then c1y1+c2y2 represent all solutions to the differential equation for all scalars c1 and c2


Homework Equations





The Attempt at a Solution


Basically, my TA's solutions to his worksheet said it was true, and I'm not sure if he implied that y1 and y2 are linearly independent solutions, since in that case it would definitely be true. I'm not sure if I got the vocab mixed up or something. (y1 can be e^x and y2 can be 2e^x and that would make my original question false, right?)
 
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freshman2013 said:

Homework Statement


If y1 and y2 are solutions to y"-y = 0 then c1y1+c2y2 represent all solutions to the differential equation for all scalars c1 and c2


Homework Equations





The Attempt at a Solution


Basically, my TA's solutions to his worksheet said it was true, and I'm not sure if he implied that y1 and y2 are linearly independent solutions, since in that case it would definitely be true. I'm not sure if I got the vocab mixed up or something. (y1 can be e^x and y2 can be 2e^x and that would make my original question false, right?)

You have it exactly right. Unless the two solutions are linearly independent the answer is "false".
 
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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