Unformatted text preview: are linearly independent on ( 1, 1).
We have g 0 (x) = 2x, while
f 0 ( x) = ( 2x,
for x > 0
2x, for x < 0 and at x = 0, we have
f (0 + h)
h f 0 (0) = lim f (0) h |h|
= lim |h| = 0.
h! 0 h
h! 0 = lim Now, W (x) = f g 0 f 0 g .
When x > 0, this is 2x3 2x3 = 0.
When x < 0, it is 2x3 + 2x3 = 0.
When x = 0, it is also 0, so the Wronskian is zero for all x 2 ( 1, 1).
Since they are linearly independent functions, Proposition 3.26 of the
Course Notes guarantees that if f and g were solutions to a 2nd-order homogeneous linear DE, then their Wronskian would not be zero. Therefore
they cannot be solutions to such an equation....
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- Winter '14
- Heterogeneity, Homogeneity, yp