Question

Hi everyone! General, Particular, Complimentary, Characteristic, Auxiliary, and whatever else I didn't think to put here. Does anyone have a way of keeping these terms from getting confusing? I am going a bit insane keeping all these straight! :o/ Thanks! :o)

Answer 1

A general solution to an ODE is one that doesn't involve initial conditions, whereas a particular solution takes them into account. Initial values gives you *particular* points that a given solution must satisfy.

Complementary solutions are typically used to denote solutions to the homogeneous version of an ODE. For instance, given the second order ODE
\[f_2(t)y''+f_1(t)y'+f_0(t)y=g(t)\]
the complementary solution is the one you get from solving
\[f_2(t)y''+f_1(t)y'+f_0(t)y=0\]
This solution (which is typically easier to find) *complements* (as in, "goes well with") the solution you get when you include \(g(t)\) on the RHS. I tend to think of the "complementary solution" as the "solution to the homogeneous part."

If the functions \(f_0,f_1,f_2\) happen to be constants, or the ODE can be rearranged so that they can be viewed as constants (e.g. via a substitution approach), then you can extract the characteristic/auxiliary equation. As far as I know, these terms can be used interchangeably, though they may mean different things in different contexts. "Auxiliary" has a connotation of meaning "secondary," in which case you can think of the auxiliary equation as a secondary equation to arise from the ODE.

Answer 2

okay, whew! I need a few minutes to absorb all this! Thanks Siths!

Answer 3

okay, so there is no such thing as a complementary equation then correct?

Answer 4

You're welcome! I would suggest having "fundamental solution" included the list. Consider the ODE
\[y'''=1\]
as a simple example. Solving for \(y\) (a matter of integrating three times), we have
\[y=\frac{1}{6}x^3+C_1x^2+C_2x+C_3\]
The fundamental solution set would be \(\{x^3,x^2,x,1\}\). What this means is that the general solution is a linear combination of these component solutions.

Answer 5

"Complementary equation" might be used to refer to the "homogeneous equation."

Answer 6

okay so when you have say y' ' + 2y'+3y + 4=0
when you start to solve it and write r^2 +2r+3(what do you do with the 4?)=0

Answer 7

do you add it to the 3?

Answer 8

do you just ignore it?

Answer 9

The auxiliary equation is a reflection of the derivatives of the dependent variable. Since \(4\) is independent of \(y\), you would have to move it to the RHS:
\[y''+2y'+3y=-4\]
so we have a *nonhomogeneous* ODE. Let's use some terminology: to get the *characteristic*/*auxiliary* equation, you would consider the *homogeneous*/*complementary equation*
\[y''+2y'+3y=0\]
which then gives the auxiliary equation
\[r^2+2r+3=0\]
We'd deal with the \(-4\) later on when solving the *nonhomogeneous equation* ODE.

Answer 10

So in a manner of speaking, you do ignore the \(4\), but you still keep track of the fact that it's a part of the ODE.

Answer 11

okay, so r^2 +2r+3=0 can be called either the characteristic equation or the auxiliary equation then?

Answer 12

Yes

Answer 13

maybe you can clear something else up for me as well...

Answer 14

when dealing with non-homogeneous DE's, three of the methods of solving them are 1)method of constant coefficients 2)annihilator approach 3) Variation of Parameters
am I missing any other methods?

Answer 15

Looks like you're dealing with second-order ODEs, is that right? There's also reduction of order. This method requires that you know one of the fundamental solutions. I can try to demonstrate an example if you like, or you can look over the exercises here: http://tutorial.math.lamar.edu/Classes/DE/ReductionofOrder.aspx

The idea is to find a solution that is linearly independent of the known solution. Variation of parameters is closely related to reduction of order.

I'm not personally familiar with the annihilator approach, at least not by that name.

Also, I think you mean method of undetermined coefficients, but it's possible that you meant what you said.

Answer 16

no you're correct...and yes 2nd order for the most part
I think that helps out a bunches!
Thanks sooo much! :o)

Answer 17

yw, good luck!