Question

Found this interesting. Have solved it, but it took me a good while.....

Answer 1

\(z_1(x)\) and \(z_2(x)\) are 2 linearly independent solutions of the DE: \(\frac{d^2 z}{dx^2} + g(x) \frac{z(x)}{2} = 0\).

show that \(\frac{d}{dx} \left( \frac{z_2}{z_1} \right) = \frac{c}{z_1^2}\) where c is a non-zero constant.

Answer 2

\[W \neq 0 \\ \text{ so } z_1 \cdot z_2'-z_2 \cdot z_1' \neq 0 \\ \\ W(z_1,z_2)(t)=ce^{- \int\limits 0 dt}=ce^{-k} \\ \text{ \choose } k=0 \\ W(z_1,z_2)=c \\ \text{ this means } z_1 \cdot z_2'-z_2 \cdot z_1'=c \text{ where } c \neq 0 \\ z_1 \cdot \frac{d z_2}{dx}-z_2 \cdot \frac{dz_1}{dx}=c \\ \\ \text{ divided both sides by } z_1^2 \neq 0 \\ \frac{1}{z_1} \cdot \frac{dz_2}{dx}-\frac{z_2}{z_1^2} \frac{dz_1}{dx}=\frac{c}{z_1^2}
\\ \frac{z_1}{z_1^2} \cdot \frac{dz_2}{dx}-\frac{z_2}{z_1^2} \frac{dz_1}{dx}=\frac{c}{z_1^2} \\ \frac{z_1 \cdot \frac{dz_2}{dx}-z_2 \cdot \frac{dz_1}{dx}}{z_1^2}=\frac{c}{z_1^2} \\ \frac{d}{dx}(\frac{z_2}{z_1})=\frac{c}{z_1^2}\]

Answer 3

I did use abel's thoerem

Answer 4

for that 3rd line

Answer 5

@freckles


this is the line i don't get, looks interesting......:
\( W(z_1,z_2)(t)=ce^{- \int\limits 0 dt}=ce^{-k} \)

Answer 6

http://ltcconline.net/greenl/courses/204/constantcoeff/linearIndependence.htm

Answer 7

so you did it without abel

Answer 8

I might have to try that

Answer 9

actually abel's theorem looks pretty easy to derive

Answer 10

never heard of Abel before! looks really interesting, i will explore.

i used algebra, basically. the Wronskian lost me hours because i couldn't convert it into a solution. the W is important, at least in my approach, to establish that \(c \ne 0\), but pretty useless otherwise.

in the attached, which is what i have submitted, you will see that i have changed the letters around a bit. so i changed \(u \rightarrow z\) and \(f(x) \rightarrow g(x)\).

[PS: if my solution is wrong, pls don't tell me !!]

Answer 11

\[z''+g \frac{z}{2}=0 \\ z_1''+g \frac{z_1}{2}=0 \\ z_2''+g \frac{z_2}{2}=0 \\ \text{ remember we have } W=z_1 \cdot z_2'-z_2 \cdot z_1' \neq 0 \\ W'=z_1 \cdot z_2''+z_1' \cdot z_2'-z_2 \cdot z_1''-z_2' \cdot z_1' =z_1 \cdot z_2''-z_2 \cdot z_1'' \\ \text{ so multiply third equation by } z_1 \\ \text{ multiply second equation by } -z_2 \\ \text{ then add the equations } \\ (z_2''+g \frac{z_2}{2})z_1+(z_1''+g \frac{z_1}{2})(-z_2)=0 \\ (z_2'' z_1-z_1''z_2)+\frac{g}{2}(z_2 z_1-z_1 z_2)=0\]
\[z_2''z_1-z_1''z_2=0 \\ W'=0 \\ W=c \\ \text{ so } z_1z_2'-z_2z_1'=c\]

Answer 12

lol

Answer 13

"[PS: if my solution is wrong, pls don't tell me !!]

"

Answer 14

brill!

Answer 15

it looks like you used what I did above

Answer 16

yes freckles, i see that too. and good to see!

[that link is really good, BTW. will read that in the light of day, ie tomorrow ... ]