Question

In solving for the solution of a non homogeneous differential equation using the method of undetermined coefficient, example, y'' + y' + y = 5. Yp will be guessed as a constant. Is that right?

Answer 1

that depends on the solution to the complimentary

Answer 2

but I guess so in this case

Answer 3

which makes finding it pretty darn easy

Answer 4

thanks :D well my problem came from the this

Answer 5

y''' - y'' = e^x + 4

Answer 6

My answer is y = yh + yp = A + Bx + Ce^x + Dxe^x + E, but my friend told me to omit the E

Answer 7

yeah, A+E is just another constant, so you can call it A again
I think your homogeneous solution is wrong though

Answer 8

I am confused, is this the same problem?

Answer 9

my above answer is the answer to the y''' - y'' = e^x + 4

Answer 10

you mean that y'' + y' + y = 5 is your answer to y''' - y'' = e^x + 4 ??
now I'm more confused

Answer 11

sorry the y'' + y' + 5 was a prelimary question. It is not related to the stuff in the comments

Answer 12

ohhhhhhhkay
so you meant that
y = yh + yp = A + Bx + Ce^x + Dxe^x is your guess to y''' - y'' = e^x + 4 ???

Answer 13

yea haha sorry for the confusion

Answer 14

now problem, but I think your guess for the particular is wrong, though I am having trouble finding it myself too

Answer 15

what did you get for yh?

Answer 16

A+ Bx+ Ce^x

Answer 17

ok, that's right (though I would call it yh=c1+c2x+c3e^x to avoid confusion)
and what did you guess for yp ?

Answer 18

Dxe^x + E , but E is a constant so same as A + E

Answer 19

that's why I said to avoid confusion, A and E are different kinds of constants. A can only be found if given an initial condition, E can be found without one

Answer 20

hmm so in this case i should not omit the E?

Answer 21

so call your guess
Axe^x+B
but that is wrong
why?
because no part of the particular solution can be linearly dependent on the complimentary
since c1 and A are both constants, we have a problem; we need to multiply by x until we get something linearly independet

Answer 22

in this case, no do not omit the E, we must make it linearly independent

Answer 23

wow thanks a lot for the explanation :D now i understand :D

Answer 24

so my answer will be y=C + C1x + C2e^x + Axe^x + Bx^2

Answer 25

oops its C1 from the start

Answer 26

almost, but rememeber that I said you can find the values of A and B

Answer 27

plug your guess for the particular into the differential equation and what do you get?

Answer 28

i need to differentiate the particular like three times and plug in

Answer 29

yep

Answer 30

hmm my question paper said there was no need to determine any coefficients

Answer 31

well, you certainly *can* find A and B if you want to, but if your paper says you don't have to find any constants at all, so be it.

Answer 32

yea, i think i could manage the finding of A and B if needed :D. Thanks for your help again :D

Answer 33

i guess ill be closing this question. Thanks for your help!

Answer 34

welcome :)