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Disagreement with my teacher about linear DE's 
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#1
Jan2513, 11:37 AM

P: 1,302

I think I may have just phrased what I meant poorly but..
If I take any linear DE, consider the independent variable (t) to be some constant, and consider y and all its derivatives to be just standard variables, and I graph this function, it always is a linear object (line, plane, etc) with dimension depending on how many y/derivatives of y are in the differential equation. I don't see how this could be false. This is a direct consequence of y and all derivatives having a linear relation, which is the definition of a linear DE. 


#2
Jan2513, 11:41 AM

P: 643

t can't be considered a constant, considering it to be a constant would cause all meaning to be lost from the derivatives.
Though, if I'm getting you correctly, you're replacing t with c (where c is an arbitrary constant) and f, f', f'', etc. with with independent variables and seeing if the resulting equation is linear wrt f, f', f'', and the other derivatives? I'm pretty sure it will be, but I don't really see why there's any reason to do this, especially since you just get meaningless jargon (again, due to the fact that the derivatives now mean nothing.) No offence, though, interesting idea! 


#3
Jan2513, 11:57 AM

P: 1,302

I know that t can't "actually" be considered constant, but for the sake of attributing meaning to why the term "linear" is used I thought of it as being "ignored" moreso than constant. He stated that in math we use words that don't always fit (linear differential equation) but with this process it fits perfectly, although I'm not very good at expressing my ideas. Like.. cos(yy') = 2t is nonlinear, and the graph of Y = arccos(C)/X = K/X Is not a line , where X is Y', C is a constant for 2t, and K is arccos(C) which is some other constant. Meanwhile 2y' + ty = 0 Is linear and Y = 2X/C Is a line with the same changes as before. 


#4
Jan2513, 06:44 PM

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P: 2,264

Disagreement with my teacher about linear DE's
2y' + ty = 0
is linear in the in the sense that if y1 and y2 are solutions c1y1+c2y2 is a solutions 


#5
Jan2513, 06:48 PM

P: 1,302




#6
Jan2513, 07:56 PM

HW Helper
P: 2,264

I do not know what you are describing. Linear DE's are linear, that is why they are so named. Perhaps you are talking about the fact that what some call affine others call linear.
If y1 and y2 are solutions then c1y1+c2y2 is a solutions is what linear means (if Ly=f instead of Ly=0 If y1 and y2 are solutions then (c1y1+c2y2)/(c1+c2) is a solutions is what linear means 


#7
Jan2513, 08:06 PM

P: 1,302

I understand that linear DEs are linear regardless of this idea (for the exact reasons you said) I am just providing an origin for what led me to think about it. I am not asking what makes a linear DE linear, I'm asking if the idea I've outlined in my first post is a true one. 


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