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I'm trying to learn DE independently, so if this is insanely stupid, 
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#19
May1111, 08:53 PM

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The complementary solution y_{c}, is the general solution to the homogeneous equation. In this case, y_{c} = c_{1} + c_{2}e^{x}. What I called the particular solution y_{p} (and you're calling f(x)) is a solution to the nonhomogenous DE, y'' + y' = x. You have f(x) = Ae^{x}. To see that it doesn't work, note that f(x) = f'(x) = f''(x) for this function, so we have Ae^{x} + Ae^{x} = x, or 2Ae^{x} = x. This has to be an identity  true for all x  and that isn't the case here. For the equation y'' + y' = x, a particular solution is y_{p} = f(x) = Ax + Bx^{2}. As for the name of the method to find y_{p}, I'm not sure there's a name. What I did was to turn the 2nd order nonhomogeneous equation y'' + y' = x into a 4th order homogeneous equation, y^{(4)} + y^{(3)} = 0, using what is called the method of annihilators. As mentioned earlier, the original DE can be represented as D(D + 1)y = x. By noting that the D^{2} operator (i.e., the second derivative with respect to x) produces 0 when applied to x, I can tack on a "factor" of D^{2} to both sides of the original DE. This gives D^{2}D(D + 1)y = D^{2}(x) = 0, which is now a homogeneous equation. The characteristic equation is r^{3}(r + 1) = 0, and the roots are r = 0 (three times) and r = 1. This gives e^{0x} and e^{1x} as solutions, but I need four independent solutions, not just two. The trick is that I can get two more functions for my set by tacking factors of x and x^{2} onto the function that's associated with the repeated roots. This gives me {e^{0}, xe^{0}, x^{2}e^{0}, e^{x}} for my set of functions, or more simply, {1, x, x^{2}, e^{x}}. Any solutions of the fourth order homogeneous DE will be a linear combination (sum of constant multiples of) these four functions. Going back to the original problem, two of these solutions (1 and e^{x}) are solutions to the homogeneous problem y'' + y' = 0. The other two (x and x^{2}) are what I picked for my "guess," looking at all linear combinations, of course. In short, y_{p} = Ax + Bx^{2}. 


#20
May1111, 09:04 PM

P: 737

LOL @ misunderstanding!!! I changed the problem so it was obvious I wasn't just copying you. That's why the first equation of the many in my last post was y''+y'=e^{x}. I hope you're not meaning (.5e^{x})''+(.5e^{x})'=.5e^{x}+.5e^{x}≠e^{x}. :)
EDIT: I forgot: Thanks, again, for your patience. 


#21
May1111, 09:24 PM

P: 737

Learning integration has taught me that heuristics, unfortunately, do have a place in math. (Referring to substitution.) Destroying my view of math as the only sanctuary of determinism, this made me sad, at the time. jk :P Great explanation, by the way. 


#22
May1211, 12:33 AM

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#23
May1211, 12:38 AM

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#24
May1211, 02:37 AM

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Looking over a selectioon of my books, you are right Mark, the dot notaion is more used for time than the prime, which is used for many other things as well.
It is, perhaps, not suprising that the Europeans should introduce the prime notation since most of their alphabets contain accented letters anyway. On another note, Tyler, did you understand my comment about integrals which was directly addressed to one of your queries/difficulties? Continuing this make sure you see the difference between [tex]\frac{{{d^2}y}}{{d{x^2}}}\quad and\quad {\left( {\frac{{dy}}{{dx}}} \right)^2}[/tex] In prime terms this is the difference between y'' and (y')^{2} go well 


#25
May1211, 01:37 PM

P: 737

I still don't know what it means to have a differential next to an expression, though... (ie what the implied function is.) 


#26
May1211, 01:58 PM

P: 737

For n repeated roots of the characteristic equation at x_{0}: y=e^{x0x}(A+Bx+...+Zx^{n1}). 


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