Question

Differential Equation:
If one fundamental solution of (t^2)y''-(3t)y'+3y=0 is y_1(t) = t, what is another fundamental solution y_2(t) satisfying y_2(1)=1 and y'_2(1)=3. (Using the definition of the Wronskian)

Answer 1

\[t^2y'' - 3ty' + 3y = 0 \]
\[y _{1}(t) = t, y _{2}(1)=1, y'_{2}(1)=3. \]
Find\[y _{2}(t)\]

Answer 2

do you know what a wronskian is?

Answer 3

wronksian, w(y_1,y_2) = y1y2' - y2y1'

Answer 4

I was thinking that y = C1t + C2e^(RT)

Answer 5

or are you at the guessing trial and error stage?

Answer 6

I'm trying to figure out the approach. So yes, im at the trial and error

Answer 7

i tried to do t^2R^2 -3tR +3 = 0, and find R in terms of T.

Answer 8

using -b+- sqrt(b^2-4ac) / 2a gives me sqrt(-3t^2) so that did not work as planned.

Answer 9

let y2 = At, were A is a function of t comes to mind ... but i cant really recall the trial and error stuff

Answer 10

what if y2 is a complex function.

Answer 11

y2 = At

y2' = A + A't , let A't = 0

y2'' = A'

want sure if you were working with complexes

Answer 12

variation of parameters is what im thinking of

Answer 13

I was working it in terms of ty2' - y2 and i know this have to be equal of a specific function..

Answer 14

or result which is not equal to zero

Answer 15

t^2 A'
- 3At -3A't
+ 3At
------------
0 = t^2 A'

if A' = 0, so let A be a constant maybe?

Answer 16

y2 = k

y1 = t

y = y1+y2

y = ct + k

Answer 17

any thoughts?

Answer 18

yes...the last method i thought of.

Answer 19

i cant say im familiar with you last method so im not able to comment on it

Answer 20

so y1 = t and y2 = e^(RT) ?

Answer 21

I'm looking up second order homogeneous diff eQ. Actually, we did not cover variation of parameters.

Answer 22

t^2 r^2 e^(rt)
- 3t r e^(rt)
+ 3 e^(rt)
-------------
0 = e^(rt) (t^2 r^2 -3tr +3)

Answer 23

this is my approach here....along with Wronskian...
http://tutorial.math.lamar.edu/Classes/DE/RealRoots.aspx

Answer 24

tr = (3 +- sqrt(-3))/2

r = (3 +- sqrt(-3))/2t

Answer 25

I found something.

Answer 26

my approach fails the conditions for y2(1) and y2'(1)

Answer 27

I think i got it.

Answer 28

does your solution work back inthe setup?

Answer 29

I used Abel's theorem and based on the equation, the W(y1,y2) = C3/t&+^3 , C3 was used a variable since integrating y2 again will bring forth another variable which is C2 and since I was given y2 and y2' initial condition i used it to substitute and find c2 and c3.
I have to try and work backwards.

Answer 30

it does not :(
the w(y1,y2) = 6/t^6 - 4/t
it should have been 2/t^3

Answer 31

it was a heck of a show tho :)

Answer 32

y2 = t^3 .... try to work towards that

Answer 33

3t^3

-3t(3t^2)

t^2 6t


3t^3 -9t^3 +6t^3 = 0

Answer 34

I attached the problem just as it is.
By the way, how did you deduce y2 to be t^3?

Answer 35

the wolf ... lol

Answer 36

what is our def of the wronskian?

Answer 37

Must not equal zero for it to have fundamental solution or be linearly independent.