Question

x^3y'' + x^2y' -4xy = 1 (x>0)
variation of parameter again
@Outkast3r09

Answer 1

\[x^3y'' + x^2y' -4xy = 1\]

Answer 2

@Outkast3r09 like i said huh ? :)

Answer 3

this one is easier

Answer 4

divide by x^3

Answer 5

realy :D

Answer 6

so it is like previous question

Answer 7

i could not do it

Answer 8

you need to figure out one solution to this equation

Answer 9

and that means

Answer 10

?

Answer 11

I figured ito ut your teacher is skipping around the book lol

Answer 12

:D

Answer 13

he probably wants you to use cauchy-eulers equation to solve for the roots

Answer 14

my book doesn't teach cauchy euler until after

Answer 15

it is also about the variation of parameter

Answer 16

at least my book says so

Answer 17

obtain a general solution using the method of the variation of parameter :/

Answer 18

yes however you need a way to find your solutions... and the only way is to using auxilary equations however in this case auxilary equations will not work

Answer 19

so you have to use cauchy-eulers method of finding the roots

Answer 20

In my book it says right here "Also, we can solve the nonhomogeneous equation ax^2y''+bxy'+cy=g(x) by variation of parameters once we have determined the yC

Answer 21

to use eulers you must have x to the exponent n derivative n

Answer 22

now i haven't taken th test on this so i'm not 100% how it works you'll have to find a derivation of this however if you let

\[y=x^m\]
\[y'=mx^{m-1}\]
\[y''=(m-1)(m)x^{m-2}\]

Answer 23

first you need the powers to line up so divide by x to get

\[x^2y''+xy'-4y=x^{-1}\]

Answer 24

then substitute in

\[x^2(m(m-1)x^{m-1}+x(m)x^{m-1}-4x^m=0\] keep it homogeneous for complementary

Answer 25

and the first is supposed to be x^{m-2}

Answer 26

\[x^{m-2}=x^mx^{-2}\] the x's cancel

Answer 27

you're left with

\[x^m(m^2-m+m-4)\]
=\[x^m(m^2-4)\]

Answer 28

\[m^2-4=0\]

m=2 k=1 , m=-2, k =1

Answer 29

since these are rational different roots you get

\[y_c=c_1x^{-2}+c_2x^2\]

Answer 30

now

\[y_1=x^{-2},y_2=x^2\]

Answer 31

\[y_p=u_1y_1+u_2y_2\]

Answer 32

make sense so far?

Answer 33

i am trying :D

Answer 34

i think that's enough for today tomorrow i will try again :) Good Night in Turkey

Answer 35

yep if you want to finish this yourself. all you have to do is take your y1 and y2 and computer the W(y1,y2) and then W1(f(x),y2) W2(y1,f(x)) and then use kramers rule u1'=W1/W