Question

Can anyone give me a step by step tutorial on how to solve an equation with an undetermined coefficient?
Ex. (D^3-D)y=2e^x
I know that I can factor out a D
(D)(D^2-1)y=2e^x and I can factor again
(D)(D+1)(D-1)y=2e^x
but I don't know what to do next.

Answer 1

@Euler271

Answer 2

well there is an exponential on the right... and that's just one.. so replace the one from e^x in the D?!??!?!

Answer 3

i have no idea what the (D - x) notation is.

the way i'd do it is: let y = C_1e^x + C_2xe^x

so (D^3 - D)y = 2e^x becomes

\[[(C_1e^x + C_2(x+3)e^x) - C_1e_x - C_2(x+1)e^x] = 2e^x\]\[2C_2e^x = 2e^x\]
so C_1 = 0 and C_2 = 1

making the particular solution: y = xe^x

hope this helps. wish i knew the notation lol

Answer 4

I think it's just a nice simple notation for the derivative operator :)\[\Large (D^3-D)y\quad=\quad y'''-y'\]

Answer 5

it's another format but it's similar to y'''+y'''+y''+y'+y=0

Answer 6

ya. i know that about it, but not looking at it in its factored form and getting info from it

Answer 7

im sure there is a simpler way to get solutions of the des just by looking at the factored form

Answer 8

if the right hand side was = 0, we would've been done, but that's not the case here

Answer 9

that is right. the total solution is the homogeneous solution [when right side = 0] + particular solution [when right side =/= 0].

sometimes the total solution is only the homogeneous part. only when = 0

Answer 10

to get the rest of the solution, you'd pretend the right side = 0

Answer 11

why am I letting y be let y = C_1e^x + C_2xe^x??

Answer 12

o...pretend that it's 0 well that would be let's see

D=0,1,-1
c1+c2e^x+c3e^-x=y

Answer 13

i think i get the notation now lol.

but for my choice of y: that is the way i'd do it and that would be a method of undetermined coefficients.

if the differential equation were D^3 - D = e^(2x), i would have chosen the form y = Ce^(2x) and played it out, giving:

8Ce^(2x) - 2e^(2x) = e^(2x) ----> C = 6 ----> y_particular = 6e^(2x)

But in our case, the right side [the particular part, aka the forcing function], taking just Ce^x leads to

Ce^x - Ce^x = 2e^x = 0

so we can't solve for C. that is why i applied the common trick of using a close form which will lead in being able to determine the coefficient(s). so i chose, by habit, C_1e^x + C_2xe^x

Answer 14

oh gawd I'm confused x(

Answer 15

so.... ok.. if maybe D = 2e^x
then