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Explain why the pair of functions $y_1(x) = x$ and $y_2(x) = \sin(x)$ cannot form a set of fundamental solutions to a second order homogeneous differential equiationon the interval $(-1,1)$.

Workings:

$W=\begin{bmatrix}x&\sin(x)\\1&\cos(x)\end{bmatrix}$

$W = x\cos(x) - \sin(x)$

If $x = 0$ then:

$W = 0\cos(0) - \sin(0)$

$W = 0$

I'm not sure if this was what i\I was supposed to do. Any help will be appreciated.

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If the Wronskian is non-zero at some initial point, then it stays non-zero at all times. See the contradiction?

$x$ satisfies $y''=0$ and $\sin(x)$ satisfies $y''+y=0$, so the minimal combination having both as solutions is $$ y^{(4)}+y''=0. $$

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  • $\begingroup$ Yeah, I do. Thanks for the help. $\endgroup$ – ineedanewnames Mar 3 '15 at 15:16
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Theorem: Supposed $f_1,\cdots,f_n$ possess at least $n-1$ derivatives. If the determinant $$\begin{vmatrix} f_1 & f_2 & \cdots & f_n\\ f'_1 & f'_2 & \cdots & f'_n\\ . &. &. &.\\ . &. &. &.\\ f_1^{(n-1)}&f_2^{(n-1)} &\cdots &f_n^{(n-1)} \end{vmatrix}$$ is not zero for at least one point in the interval I, then these functions are linearly independent on the interval.

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  • $\begingroup$ Could you explain the connection to the question? Are the $f_k$ solutions of a linear ODE? Is linear independence meant as functions on the interval, then the statement is trivial, or pointwise for the the derivative vectors, then it is wrong. Also note that $x$ and $\sin x$ are linearly independent as functions. $\endgroup$ – Lutz Lehmann Mar 3 '15 at 15:27

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