Question

Show by Theorem 6 that the functions in (H) are not solutions on [-1,1] of any second order homogeneous linear equation.

Theorem 6 states that If y1,...,yn are n solutions of the nth order homogeneous equation y^(n)+Pn-1y^(n-1)+...+p1y'+poy=0 and W(y1(xo),...,yn(xo))=0 for some xo in (a,b), then y1,...,yn are linearly depended on (a,b) and hence W(y1(x),...,yn(x)=0 on (a,b))

The functions in H are 1 and cos x
y = 1 y' =0 y'=0
y = cosx y'= -sinx
taking the determinant, the result is -sinx

Answer 1

I was wondering if I should take the values of -1 and 1 into sin x for consideration since sinx = 1 occurs when it's pi/2 and sinx = -1 occurs when it's 3pi/2

Answer 2

ugh it's supposed to be a 2x2 matrix for the Worsk...ahhh let's call it W
y = 1 y' =0
y = cosx y'= -sinx

Answer 3

and then the determinant is -sinx

Answer 4

@hartnn

Answer 5

n solutions... 1 solution 2 solution 3 Mississippi

Answer 6

@bahrom7893

Answer 7

X(

Answer 8

I suppose you want to find \(x\) values where \(W=0\)?

Answer 9

What do you think the relationship between linear dependents and and homogeneous equations are?

Answer 10

linearly dependent... there is some value but not all are 0

Answer 11

like v1+4v2+3v3=0
I do know that linearly independent is c1=c2=c3=0

Answer 12

but isn't like if I take sin 3pi/2 wouldn't that be 0?

Answer 13

well it does look similar...are they related? C>C

Answer 14

hmm if I wanted zero values... can't -sinx be 0, pi, and 2pi?

Answer 15

Since we are only considering linear homogeneous 2nd order differential equations...

Answer 16

We're only considering \[
y''+ay'+b=0
\]

Answer 17

which means that the function is at least y double prime

Answer 18

so do I take -sinx and take the derivative twice?

Answer 19

-cos x = y'
sinx = y"