Question

Non-homogeneous linear differential equation question:

Well my professor is quiet terrible, constantly equivocating and over complicating problems, so I turned to the Khan Academy. For non homogeneous equations how do you know how many variables to use to represent the right hand section of the equation? For example...
y''-3y'-4y=4x^2
Now I know how to solve the homogenous part of this equation, but the 4x^2 is where I get lost.
How do I know if I should us Ax^2+bx+C or just an _A+_B or just a single variable to represent it?

Answer 1

the way i learnt to tackle problems like this is through variations of parameters.
so solving the characteristic equation, you get the roots which correspond to a specific solution with arbitrary constants in front. i.e. for example y = Ae^4x + Be^-x
now you take this solution, and differentiate it twice, keeping your terms for y' and y along the way. now your original equation can be written as a linear combination of these terms. (it's important along the way to treat A and B as functions not constants, and it is useful to invoke restrictions such as:
y = A e^4x + B e^-x
y' = 4Ae^4x - Be^-x + A'e^4x + B'e^-x {let 1) A'e^4x + B'e^-x = 0} <--)
you'll be left with a simple linear system with only first derivatives, you can then find functions A and B.
I know i've just flow through this, but if you'd like to read more:
http://www.cliffsnotes.com/study_guide/Variation-of-Parameters.topicArticleId-19736,articleId-19722.html

Answer 2

So from what I'm getting at, it seems as though it is some what random how many function variables such as A, B, ad C there are?

Answer 3

when solving an nth order equation, generally there will always be n arbitrary constants. i.e. in 2nd order you'll be solving for only 2 unknown functions A and B

Answer 4

That's what I thought as well, but in the equation y''-3y'-4y=4x^2 (www.khanacademy.org/math/differential-equations/v/undetermined-coefficients-3) he set 4x^2 Yp=Ax^2 + Bx + C as opposed only two variable functions.

Answer 5

@JamesJ @robobey @lgbasallote @myininaya @mertsj anyone??

Answer 6

are you sure that Ae^4x + Be^-x + C is a solution to y'' - 3y' -4y = 0 ?

Answer 7

The homogeneous part of the equation would be
C1(e^4x) + C2(e^-x)
I know that part just not how to deduce how many variables such as A or B there will be. I thought it was dependent on the order, but apparently not according to this video

Answer 8

yeah, so to do variation of parameters you just find the two functions C1 and C2 from the solution of the homogenous equation. there should only be as many variable as the number of them that appear in solving the homogenous equation (in this case 2)

Answer 9

So the C is arbirtrary in this as it becomes zero with the first derivative?

Answer 10

if you were to include C you'd end up with 2 equations for 3 functions and i'm not sure it would work out (or maybe it does i haven't tried it out) but what i do know is that if you stick with the homogenous solution and solve for the constants in that you'll always get a consistent solution

Answer 11

Now it is starting to make sense. Thank you, much appreciated!

Answer 12

anytime!