Question

Find the general solution to the follow differential equation:
f′′ − 6f′ + 9f = 20 − 12x + 9x^2.

Answer 1

i understand how to get the e part i just need help with the end part!

Answer 2

whats you characteristic equation to solve for your fs

Answer 3

oh, the end can be tabled

Answer 4

yp = ax^2 +bx +c
and take derivatives and plug them into your diffyQ

Answer 5

well, fp for this since they wanna be weird and all :)

Answer 6

ahah wait what!

Answer 7

what derivitives do i plug into where?

Answer 8

the derivatives of the generic quadratic that matches up to the specific quadratic

Answer 9

fp = ax^2 + bx + c is a generic match to 9x^2 -12x +20

Answer 10

find fp' and fp'' and plug then into the diffQ to compare parts with

Answer 11

so fp' = 18x - 12, fp'' = 18?

Answer 12

fp′′ − 6fp′ + 9fp = 9x^2-12x+20

fp = ax^2 + bx + c

fp' = 2ax + b

fp'' = 2a

Answer 13

fp′′ − 6fp′ + 9fp = 9x^2-12x+20

\[2a -6(2ax + b) +9(ax^2 + bx + c) = 9x^2-12x+20\]

Answer 14

ohhh ok.

Answer 15

then do i factor an x out of like termS? or?

Answer 16

expand it all; gather up your like terms; and compare them to their respective parts from the right side

Answer 17

9ax^2 = 9x^2; a=1

that one an easy one

Answer 18

so a = 1, b = -4/3, c=20/9 ?

Answer 19

9bx + 12ax = -12x

(9b +12)x = -12x

9b = -24; b=-24/9 = -12/6 = -4/3

yeah, stuff like that

Answer 20

fp = ax^2 +bx + c

we found a,b and c

fp = x^2 -4x/3 +20/9

Answer 21

you say you found the homogenous part; so add them up

f = fh + fp

Answer 22

ok so f(x) = Ae^3x + Bxe^3x + x^2 -4x/3 +20/9

Answer 23

yep, as long as your fh is good ;)

Answer 24

there might be an issue with your x terms tho; i recall someting about something that i forgot a while ago that might make that one an issue

Answer 25

uh oh, whys that!

Answer 26

something to do with keeping all the parts linearly independant

Answer 27

http://www.wolframalpha.com/input/?i=y%27%27-6y%27%2B9y%3D9x%5E2-12x%2B20

yeah, looks like that might be an issue since the solutions dont match up for us

Answer 28

oh no so what did i do wrong ahha

Answer 29

try it again but we have to bump up the xparts;

x^2(ax^2+bx+c)

ax^4+bx^3+cx^2

would seem to me to cover it; but thats just me trying to remember this

Answer 30

hmmm ok

Answer 31

i dont think i'm getting it still ..

Answer 32

i checked the wolfs steps and they did what we did; i think i typed it in right :)

double check the maths from the start of it

Answer 33

2a
-12ax -6b
9ax^2+9bx+9c
--------------
9x^2−12x+20

a=1


2
-12x -6b
9x^2+9bx+9c
--------------
9x^2−12x+20

-12+9b=-12 ; found it ... b=0



2
-12x -
9x^2+ +9c
--------------
9x^2−12x+20


2+9c = 20 ; c=2

Answer 34

fp = x^2 + 2 :D