KILLER 2nd ODE (inhomogeneous) XD

  • Thread starter Thread starter Zomboy
  • Start date Start date
  • Tags Tags
    Ode
Zomboy
Messages
6
Reaction score
0
Ok, here goes:

Homework Statement

So I've come across this 2nd ODE which I am to "solve ... for a general solution":

d^2y / dx^2 - dy/dx + y = cos(x) - sin(x) :-p

and then evaluate the "particular solution" using the boundary conditions y=L when x=0 (also, dy/dx = 0)

The Attempt at a Solution



I can't type out the whole of my working because its really long and would be impossible to follow so I'll try and sum up what I've got:
1) found the general solution of the * equivalent* homogeneous equation... which came out with imaginary values. I then converted this into trigonometric form (as opposed to using imaginary exponentials) which is in the form of:

exp(1/2 x) ( C sin((sqrt(3)/2)x) + D cos("") )
2) I then guessed at the particular solution which I'm thinking looks like:

(a-b)( cos(x) - sin (x) ) :rolleyes:
3) Added ^^these^^ together to get the "General Solution" (y=...) of the original equation. Which looks something like (but with more coefficients and stuff:

e^... (cos + sin) + ( cos - sin ) {you get the idea} :bugeye:4) Trying to evaluate this however lead me to some nasty unsolvable simultaneous equations...

any advice? can you spot my mistake? do I actually just need to solve "2)" to get the answer, I'm confused now...

this is driving me absolutely crazy.
 
Physics news on Phys.org
The particular solution should be of the form:
[tex] Y_p(x) = A \, \cos x + B \, \sin x[/tex]

Determine A and B by plugging this in the ODE. Tell us what you get.
 
Zomboy said:
Ok, here goes:

Homework Statement




So I've come across this 2nd ODE which I am to "solve ... for a general solution":

d^2y / dx^2 - dy/dx + y = cos(x) - sin(x) :-p

and then evaluate the "particular solution" using the boundary conditions y=L when x=0 (also, dy/dx = 0)




The Attempt at a Solution



I can't type out the whole of my working because its really long and would be impossible to follow so I'll try and sum up what I've got:



1) found the general solution of the * equivalent* homogeneous equation... which came out with imaginary values. I then converted this into trigonometric form (as opposed to using imaginary exponentials) which is in the form of:

exp(1/2 x) ( C sin((sqrt(3)/2)x) + D cos("") )



2) I then guessed at the particular solution which I'm thinking looks like:

(a-b)( cos(x) - sin (x) ) :rolleyes:



3) Added ^^these^^ together to get the "General Solution" (y=...) of the original equation. Which looks something like (but with more coefficients and stuff:

e^... (cos + sin) + ( cos - sin ) {you get the idea} :bugeye:





4) Trying to evaluate this however lead me to some nasty unsolvable simultaneous equations...

any advice? can you spot my mistake? do I actually just need to solve "2)" to get the answer, I'm confused now...

this is driving me absolutely crazy.

Suggestions:
(1) Use a Green's function method http://www.math.umn.edu/~olver/pd_/gf.pdf ; or
(2) Use Variation of Parameters http://en.wikipedia.org/wiki/Variation_of_parameters

Both methods are standard, but (2) is probably better known.

RGV
 
Last edited by a moderator:

Similar threads

Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 11 ·
Replies
11
Views
4K
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
3
Views
3K