# I'm trying to learn DE independently, so if this is insanely stupid,

by TylerH
Tags: independently, insanely, learn, stupid
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 Quote by TylerH After doing a little research on him, I have a theory on why his notation is used so prevalently in diff eq. He succeeded Euler, who discovered characteristic equations and did lots of other founding work on DE (Euler method of numerical solving comes to mind), at the Prussian Academy of Sciences. One can guess that, with all the discoveries being made by Euler at the time, the Prussian Academy of Sciences probably became the premier DE school. It also probably retained it's status when Lagrange took over. He probably pushed the usage of his notation. Not that I'm saying all he did was piggy back off the immense awesomeness of Euler. Apparently, he did some serious number theory stuff. Now, to prove I'm no dumby head: I think I've figured it out: y''+y'=ex y=erx+f(x) You called f(x) "yp" r2+r=0 r={-1,0} y=C+De-x+f(x) f(x)=Aex Not really sure how to justify this. It just came to me. (C+De-x+Aex)''+(C+De-x+Aex)'=ex solves to A=1/2 y=C+De-x+.5ex EDIT: Should I leave C and D or let them be 0? Is there a name for the nonguess and check technique used for finding f(x)/yp?
This won't work, and you can confirm that it doesn't work by substituting your f(x) into the nonhomogeneous DE.

The complementary solution yc, is the general solution to the homogeneous equation. In this case, yc = c1 + c2e-x.

What I called the particular solution yp (and you're calling f(x)) is a solution to the nonhomogenous DE, y'' + y' = x.

You have f(x) = Aex. To see that it doesn't work, note that f(x) = f'(x) = f''(x) for this function, so we have Aex + Aex = x, or 2Aex = 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 yp = f(x) = Ax + Bx2.

As for the name of the method to find yp, 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 D2 operator (i.e., the second derivative with respect to x) produces 0 when applied to x, I can tack on a "factor" of D2 to both sides of the original DE.

This gives D2D(D + 1)y = D2(x) = 0, which is now a homogeneous equation.

The characteristic equation is r3(r + 1) = 0, and the roots are r = 0 (three times) and r = -1.

This gives e0x 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 x2 onto the function that's associated with the repeated roots.

This gives me {e0, xe0, x2e0, e-x} for my set of functions, or more simply, {1, x, x2, 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 x2) are what I picked for my "guess," looking at all linear combinations, of course. In short, yp = Ax + Bx2.
 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'=ex. I hope you're not meaning (.5ex)''+(.5ex)'=.5ex+.5ex≠ex. :) EDIT: I forgot: Thanks, again, for your patience.
P: 737
 Quote by Mark44 This gives e0x 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 x2 onto the function that's associated with the repeated roots. This gives me {e0, xe0, x2e0, e-x} for my set of functions, or more simply, {1, x, x2, e-x}.
How did know you choose x and x2? Heuristics is a fine answer if there is no other explanation.

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.
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 Quote by TylerH 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'=ex. I hope you're not meaning (.5ex)''+(.5ex)'=.5ex+.5ex≠ex. :) EDIT: I forgot: Thanks, again, for your patience.
I didn't realize that you had changed the right side on the DE.
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 Quote by TylerH How did know you choose x and x2? Heuristics is a fine answer if there is no other explanation.
I'm not sure what the origin of this is. There's a lot of mathematics that has arisen because someone discovered a useful technique.
 Quote by TylerH 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.
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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?

 I'm finding it hard to understand what exactly is means to have a differential after an expression ie "2xdx."
I am trying to supplement, not supplant, the excellent job Mark is doing at explanation here, but I agree with AZ that you have some of the basics muddled up.

Continuing this make sure you see the difference between

$$\frac{{{d^2}y}}{{d{x^2}}}\quad and\quad {\left( {\frac{{dy}}{{dx}}} \right)^2}$$

In prime terms this is the difference between y'' and (y')2

go well
P: 737
 Quote by Studiot On another note, Tyler, did you understand my comment about integrals which was directly addressed to one of your queries/difficulties? I am trying to supplement, not supplant, the excellent job Mark is doing at explanation here, but I agree with AZ that you have some of the basics muddled up. Continuing this make sure you see the difference between $$\frac{{{d^2}y}}{{d{x^2}}}\quad and\quad {\left( {\frac{{dy}}{{dx}}} \right)^2}$$ In prime terms this is the difference between y'' and (y')2 go well
Yes, I believe I understood. The thanks were directed at all, not just Mark. :)

I still don't know what it means to have a differential next to an expression, though... (ie what the implied function is.)
P: 737
 Quote by Mark44 I'm not sure what the origin of this is. There's a lot of mathematics that has arisen because someone discovered a useful technique.
I did some research. http://en.wikipedia.org/wiki/Charact...ted_real_roots

For n repeated roots of the characteristic equation at x0: y=ex0x(A+Bx+...+Zxn-1).