Question

variation of parameters

Answer 1

#6

Answer 2

just input those bad boys on wolfy alphy and print screen k? you can do both of them :)

Answer 3

because my brain is tired that's why

Answer 4

oh I did 9 and 10 already... so yeah just 6 and then I have to do 1b 1c got part of 1b

Answer 5

lol I don't understand, you don't have access to wolfram? :3
Too lazy?

Answer 6

no I used my three daily solutions...can you please input these things

Answer 7

Oh I see :3

Answer 8

yeah I reached my limit by accident

Answer 9

if you could just plug those two equations into wolfram and then print screen the details via variation of parameters, it would mean a lot.

Answer 10

k simmer down :) gimme a sec lol

Answer 11

thank you :D you'll get a medal for your work

Answer 12

y''+(1x^-1)y'-(1x^-2)y=0
http://www.wolframalpha.com/input/?i=y%27%27%2B%281x^-1%29y%27-%281x^-2%29y%3D0
y''+(1x^-1)y'-(1x^-2)y=lnx
http://www.wolframalpha.com/input/?i=y%27%27%2B%281x^-1%29y%27-%281x^-2%29y%3Dlnx
click on step by step solution and variation of parameters...if that's not there, undetermined coefficents. either one is fine

Answer 13

Oh it looks like the homogeneous case is solved within the second one :o
(I guess it would have to be)

Answer 14

umm there's more to that... the Yp part

Answer 15

@zepdrix oy there's more than the Yh...need the Yp

Answer 16

that's just a part of it bro

Answer 17

go further down and print screeen plzzzz

Answer 18

@Euler271 I need a favor plz

Answer 19

y''+(1x^-1)y'-(1x^-2)y=lnx http://www.wolframalpha.com/input/?i=y%27%27%2B%281x^-1%29y%27-%281x^-2%29y%3Dlnx click on step by step solution and variation of parameters...if that's not there, undetermined coefficents. either one is fine

Answer 20

I've used my limits by accident -_-

Answer 21

if you can print screen the solution, that would be great...there's more than what zepdrix put

Answer 22

lol i hate when that happens

Answer 23

did you use all of yours Euler?

Answer 24

zepdrix that's just Yh I need the whole thing up to the Answer: | | y(x) = y_c(x) + y_p(x) = c_1/x+c_2 x+1/9 x^2 (3 log(x)-4)

Answer 25

no i have

Answer 26

plz help me D:

Answer 27

medals for all in this thread

Answer 28

Hmm weird.. there was one final step beyond what I put, but Wolfram is tweaking out on me for some reason :(

When I click "Show all" it turns black...
I can't get the final step for some reason D:

Answer 29

Euler save us!

Answer 30

tweaking or twerking?

Answer 31

lol :3

Answer 32

save me D:

Answer 33

uploading ^_^

Answer 34

4 s/s:

1: http://imageshack.us/f/14/9e7y.jpg/
2: http://imageshack.us/f/69/6c2k.jpg/
3: http://imageshack.us/f/820/vym5.jpg/
4: http://imageshack.us/f/36/ls55.jpg/

Answer 35

I can't see it D:

Answer 36

too small

Answer 37

lol take these:

http://imageshack.us/a/img14/632/9e7y.jpg
http://imageshack.us/a/img69/527/6c2k.jpg
http://imageshack.us/a/img820/7153/vym5.jpg
http://imageshack.us/a/img36/3575/ls55.jpg

Answer 38

you can also click on them to make them bigger [previous links]

Answer 39

thank you! :D

Answer 40

oh crud one medal per person D:

Answer 41

Euler give a medal to zep and zep give to me

Answer 42

lol np ^_^

Answer 43

oh you got the solution? c: yay

Answer 44

yeah but how do I prove that x and 1/x are solutions to y''+(1x^-1)y'-(1x^-2)y=0

Answer 45

@dumbcow

Answer 46

wow id rather not read all the posts...ok so you have the solution and want to verify solutions right?

Answer 47

yes

Answer 48

prove that x and 1/x are solutions to y''+(1x^-1)y'-(1x^-2)y=0

Answer 49

that part...I did the solving already

Answer 50

sorry I was in zombie state I mean normally I would solve and verify with wolfram but I was stressing out

Answer 51

so where do I begin?

Answer 52

could be related to this
http://tutorial.math.lamar.edu/Classes/DE/VariationofParameters.aspx

Answer 53

no stress , solving is hard part, to verify just find y' and y'' then plug it in and it should equal 0

y = x
y' = 1
y'' = 0

y = 1/x
y' = -1/x^2
y'' = 2/x^3

Answer 54

love you :D

Answer 55

yes just one more problem to go! woo!