Question

For the interval x > 0 and the function set
S = { 3ln(x), ln2, ln(x), ln(5x)}
construct a linear ODE of the lowest order

Answer 1

After taking the wronskian, I get W = 0. Doesn't that mean that there is no linear ODE for this set?

Answer 2

yeah, it would I believe...

Answer 3

one of your solutions is ln2 ???

Answer 4

yeah ln2. I'm not quite certain if I should be excluding that or not, since it can only be differentiated once.

Answer 5

I don't think that wronskian is zero
how did you take the determinant of the 4x4 matrix? method of cofactors?

Answer 6

oh wait, lnx and 3lnx are linearly dependent, so...

Answer 7

well the ln2 column gives 3 zeros, right?
So I just multiplied -ln2 with the remaining 3x3 matrix

The 3x3 matrix is a wronskian of {3/x,1/x,1/(5x)} --- (the derivates of the solutions)

Does that sound right so far?

Answer 8

yeah that is method of cofactors

Answer 9

let me see if I can get the solution up here

Answer 10

d/dx(ln(5x))=1/x

Answer 11

careful with the chain rule there...

Answer 12

aaahhh that's right! Can't believe I messed that up

Answer 13

so I get\[W=\left|\begin{matrix}3\ln x&\ln x&\ln(5x)\\\frac3x&\frac1x&\frac1x\\-\frac3{x^2}&-\frac1{x^2}&-\frac1{x^2}\end{matrix}\right|\]looks fishy to me

Answer 14

rather\[W=-\ln2\left|\begin{matrix}3\ln x&\ln x&\ln(5x)\\\frac3x&\frac1x&\frac1x\\-\frac3{x^2}&-\frac1{x^2}&-\frac1{x^2}\end{matrix}\right|\]

Answer 15

Won't you have the first row of the new matrix as | 3/x 1/x 1/x | ?
Since the original first row with 3lnx etc is multiplied by 0?

Answer 16

oh shnap, that's why two heads are better than one

Answer 17

\[W=-\ln2\left|\begin{matrix}\frac3x&\frac1x&\frac1x\\-\frac3{x^2}&-\frac1{x^2}&-\frac1{x^2}\\\frac9{x^3}&\frac3{x^3}&\frac3{x^3}\end{matrix}\right|\]

Answer 18

Isn't the derivative of -1/x2 = 2/x3 ?

Answer 19

\[W=-\ln2\left|\begin{matrix}\frac3x&\frac1x&\frac1x\\-\frac3{x^2}&-\frac1{x^2}&-\frac1{x^2}\\\frac6{x^3}&\frac2{x^3}&\frac2{x^3}\end{matrix}\right|\]you know, I really hate arithmetic

Answer 20

haha who doesn't

Answer 21

that has to be zero because the last two columns are the same

Answer 22

Yeah I get the same

Answer 23

(I'm not actually bothering to do the calculation)

Answer 24

yeah don't worry about that, the determinant is 0

Answer 25

It seems like the original set has to be rewritten

Answer 26

the problem seems to stem from the fact that all the derivatives are linearly dependent constant multiples of each other, (since the constants cancel due to the chain rule) so this thing seems not to want to be a fundamental set of solutions for anything

Answer 27

it's weird to have one of the solutions be a constant too, I just don't know what to make of that

Answer 28

Yeah I'm discussing this on another forum at the moment. Someone suggest that I would need to take the smaller set of {ln 2, ln x}