Question

Suppose that L(y)= ...

Answer 1

\[L \left[ y \right] = (x ^{2}+1)y \prime \prime-2xy \prime+2y = (x ^{2}+1)^{2}\]

and \[y _{1}= x ^{2}+1\],\[y _{2} = x\] are solutions of L[y] = 0. Find the general solution of \[L \left[ y \right] = (x^{2}+1)^{2}\] on the interval .

Answer 2

interval \[\left| x \right|<1\]

Answer 3

is L a Laplace transform?

Answer 4

No.

Answer 5

then youll have to define it for me. The only other time Ive seen it would be as a linear operator then

Answer 6

yes, that would be it.

Answer 7

hmmm, so it looks like y = y1 + y2 + y3 + ... is what it is refereing to

Answer 8

yes.

Answer 9

variation of parameters?

Answer 10

quite possibly, using a wronskian perhaps?

Answer 11

i know the long version, and the short version ....

Answer 12

that too.

Answer 13

the short version involve the theorem?

Answer 14

lol, i never recall thrms. The wronskian is a determinant of derivatives i beleve

Answer 15

yup

Answer 16

let me know the short version for now.

Answer 17

\[\begin{vmatrix}W_x&W&W_y\\
y_1&0&y_2\\
y'_1&g(x)&y'_2\\
\end{vmatrix}\]

Answer 18

this actually ends up being the cramer rule for a solution of 2 equation in 2 unknowns if we do the long version

Answer 19

do you know how to run a determinant on this matrix setup?

Answer 20

the ones i have dealt with only are 2 by 2, so not really.

Answer 21

this setup takes advantage of submatrixes

to find Wx cover the row/column of the Wx and take the determinant of whats left: -y2 * g(x)

to find Wy cover the row/column of the Wy and take the determinant of whats left: y'1 * g(x)

to find W cover the row/column of the W and take the determinant of whats left: y1 y'2 - y2 y'1

Answer 22

it tends to be more intuitive if taken the long way first .... but im a little slow :)

Answer 23

i would disagree.

Answer 24

with the 2nd phrase.

Answer 25

lol

Answer 26

i would like to be as slow as you are hehe

Answer 27

could you go step by step with me. i think im really slow.

Answer 28

im trying to recall the rest of the setup using the wronskian

i know its something to do with Wx/W and so forth

Answer 29

yes, its pretty tedious stuff.

Answer 30

Please dont hesistate to bring in other people. Thank you very much!

Answer 31

isnt the Wonskian this way
\[
W(t)=\left[\begin{matrix}x^2+1&x\\2x&1
\end{matrix}\right]
\]

Answer 32

i was thinking about that..

Answer 33

the smarter than mes tend to come in and peek about :)

\[
\begin{vmatrix}W_x&W&W_y\\
x^2+1&0&x\\
2x&(x^2+1)^2&1\\
\end{vmatrix}\]
\[W_x=-x(x^2+1)^2\\
W = (x^2+1)-2x^2\\
W_y=(x^2+1)^3
\]

\[\frac{W_x}{W}=\frac{-x(x^2+1)^2}{1-x^2}=\frac{-x(x^2+1)^2}{-(x^2-1)}\]

Answer 34

I do a setup that allows me to just run all the Ws at a glance

Answer 35

I see.

Answer 36

Thats the Cramer's Rule for solution of set of equations?

Answer 37

yes

Answer 38

\[A'x+B'y=0\\A'x+B'y=g(x)\]

Answer 39

you guys are too smart for me haha

Answer 40

I got the Qronksian down, what next?

Answer 41

Wronskian*

Answer 42

wow! a new method, the Qronskian :)

Answer 43

hey!

Answer 44

hahaha

Answer 45

@qrious126 this explains the same. you want to check out
http://scipp.ucsc.edu/~haber/ph116C/Wronskian_12.pdf

Answer 46

math is the only language that advances "for" science based on old math. Other languages develop for no reasons!!

Answer 47

hahah

Answer 48

you know you can learn so much if you dont proscrastinate almost every time you're given an assignment.

Answer 49

the long version is .....

yp = A(x) y1 + B(x) y2

y'p = A(x) y'1 + B(x) y'2 + A'(x) y1 + B'(x) y2 ; let A'(x) y1 + B'(x) y2 = 0 for the homogenous part, giving us:

y'p = A(x) y'1 + B(x) y'2

y''p = A(x) y''1 + B(x) y''2 + A'(x) y'1 + B'(x) y'2

inserting the yps into the original equation gives us

A(x) y''1 (x^2+1) + B(x) y''2 (x^2+1) + A'(x) y'1 (x^2+1) + B'(x) y'2 (x^2+1)
-2x A(x) y'1 - 2x B(x) y'2
+2 A(x) y1 + 2 B(x) y2
----------------------
(x^2+1)^2

which should give us 2 equations in A' and B' to solve

Answer 50

thanks! appreciate your help!

Answer 51

@amistre64 what did I do to deserve a medal?

Answer 52

you helped me out with that link you posted :)

Answer 53

@amistre64 hmm. thanks. and welcome ^_^