Question

Solve by variation of parameters

Answer 1

\[xy''-4y'=x^4\]

Answer 2

\[x^2y''-4xy'=x^5\]

Answer 3

\[m^2-m-4m=0\]
\[m(m-5)=0\]
m=0 , m=5

Answer 4

\[y_c=c_1+c_2x^5\]

Answer 5

\[W(y_1,y_2)=1(5x^4)-0=5x^4\]


\[y_p=-y_1\int \frac{x^5(x^5)}{5x^4}+y_2\int \frac{x^5}{5x^4}\]

Answer 6

@eliassaab

Answer 7

does this look correct?

Answer 8

cause i swear the answer in the book has an ln(x) but i cannot figure out wy in the world it would

Answer 9

@TuringTest

Answer 10

You see anyway this can be ln(x)

Answer 11

@jim_thompson5910

Answer 12

Does this have to do somewhat with x^5 being a solution to the complementary

Answer 13

hm... *scratching head*

Answer 14

I think variation of parameters is just for constant coefficients

Answer 15

that's what I was thinking, but the problem says...
it wouldn't be the first time I've seen someone use a technique that I thought inapplicable to some form that turned out to work

Answer 16

But you should check your solution to the complimentary solution. I just did constant coefficients in an Engineering Analysis in grad school and what your doing doesn't line up with that at all. Check out this guys notes. I used them alot http://tutorial.math.lamar.edu/Classes/DE/HOVariationOfParam.aspx

Answer 17

the complimentary is correct

Answer 18

http://www.wolframalpha.com/input/?i=x*y%27%27-4y%27%3Dx%5E4
just so we know what we're aiming for^

Answer 19

well this is going to keep me up tonight...

Answer 20

what lol

Answer 21

oh I just mean it's going to drive me nuts trying to find out how to get the right solution is all

Answer 22

oh stupid me

Answer 23

i know what it s

Answer 24

i didn't put it in high order linear form

Answer 25

please enlighten me

Answer 26

in order to use Variation of parameters it needs to be in form

y''+p(x)y'+q(x)y=f(x)

Answer 27

i have to divide by x^2 before doing anything

Answer 28

oh, so y'' has to have a coefficient of 1 !!
that's right, I do remember something like that now...

Answer 29

but then what is the characteristic polynomial?

Answer 30

...must be the same I guess

Answer 31

no characteristic

Answer 32

using variation of parameter divide by x^2 to get the f(x0

Answer 33

once you have x you just use y1, y2 to find the wronkskian

Answer 34

but the coefficient of y' isn't constant, how do you deal with that?

Answer 35

\[y_p=-y_1\int \frac{x^3*x^5}{5x^4}+y_2\int\frac{x^3}{5x^4}\]

Answer 36

\[\frac{x^8}{5x^4} =\frac{1}{5}x^4\]

Answer 37

taking the integral you get

\[\frac{1}{25}x^5\]

Answer 38

oh yeah, you're gonna get the right answer all right...
but answer my question! your methods are different from mine and I want to understand them
how do you deal with the non-constant coefficient of y' ?

Answer 39

i don't believe it matters what the coefficent is

Answer 40

seen as though you are only trying to get the f(x) function

Answer 41

then how did you get the complimentary?

Answer 42

Euler-Cauchy's Equation

Answer 43

that says sub. y=x^m

Answer 44

this you can only use when the power of x is the same as the prime itself

Answer 45

^that was what I didn't know
I will be researching that. Thanks!