Linear Independence, Differential Equations

[Ted123]

Homework Statement

[PLAIN]http://img220.imageshack.us/img220/7427/diff5.jpg

The Attempt at a Solution

Done (a). How do I go about (b) and (c)?

 

[Mark44]

For b, to construct the equation y'' + y = 0, you must have p(x) \equiv 0 and q(x) \equiv 1. What implication does this have on what you established in part a?

 

[Ted123]

For b, to construct the equation y'' + y = 0, you must have p(x) \equiv 0 and q(x) \equiv 1. What implication does this have on what you established in part a?

y_1 y_2^{''} = y_2 y_1^{''}

and

W(y_1,y_2) = y_1^{'} y_2^{''} - y_2^{'} y_1^{''}

 

[Mark44]

You also know that {sin x, cos x} and {sin x, sin x - cos x} are two pairs of solutions of y'' + p(x) y' + q(x) y = 0.

 

[Ted123]

So for the first set of solutions we just let y_1 = \sin x and y_2 = \cos x.

\Rightarrow p(x) = -\frac{-\sin x \cos x + \cos x \sin x}{W(y_1,y_2)} = 0

and q(x) = \frac{- \cos ^2 - \sin^2 x}{W(y_1,y_2)}

and W(y_1,y_2) = -\cos^2 x - sin^2x

So subbing p and q in:

y'' + \frac{-\cos ^2 x - \sin^2 x}{-\cos^2 x - sin^2x} y = y'' + y \stackrel{!}{=} 0.

And the same for the other solutions?

 

[Mark44]

Yes, try it and see what you get.

BTW,
-\cos^2 x - sin^2x = -(\cos^2 x + sin^2x ) = -1