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Let $u$ and $v$ be two solutions of the differential Equation

$y^{"} + P(x)y^{'} + Q(x)y = 0$ on $[a,b]$, Let $W(u,v)$ denote the Wronskian Of $u$ and $v$ Then

(a) $W(u,v)$ vanishes at point $x_{0} \in [a,b]$ $\implies$ $u,v$ are Linearly Dependent.

(b) $W(u,v)$ is identically zero on $[a,b]$ $\implies$ $u,v$ are Linearly Dependent

Now ,I Know from this question Proof that ODE solutions with Wronskian identically zero are linearly dependent That option (b) must be correct.

For option (a) We know that Wronskian for Differential Equation is either identically Zero or Never Zero, so if Wronskian vanishes at one point $x_0 \in [a,b]$ it must vanish identically in $[a,b]$ so option (a) is also correct.

So both options (a) and (b) must be true for this question.

Is My answer correct ?

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1 Answer 1

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From the definition

$W(u, v) = \det \begin{bmatrix} u & v \\ u' & v' \end{bmatrix} = uv' - vu' \tag 1$

we find

$W'(u, v) = u'v' + uv'' - v'u' - vu'' = uv'' - vu''; \tag 2$

using the fact that $u$ and $v$ are both solutions of

$y'' + P(x)y' + Q(x)y = 0, \tag 3$

we may transform (2) to

$W'(u, v) = u(-P(x)v' - Q(x)v) - v(-P(x)u' - Q(x)u) = -P(x)uv' - Q(x)uv + P(x)vu' + Q(x)uv = -P(x)(uv' - vu') = -P(x)W(u, v); \tag 4$

the solution to this ordinary linear equation for $W(u, v)$ is readily seen to be

$W(u, v)(x) = \exp \left (-\displaystyle \int_{x_0}^x P(s) \; ds \right )W(u, v)(x_0); \tag 5$

under fairly mild conditions on $P(x)$ we may affirm that

$\forall x, \; \exp \left (-\displaystyle \int_{x_0}^x P(s) \; ds \right ) \ne 0, \tag 6$

and thus

$W(u, v)(x_0) \ne 0 \Longrightarrow \forall x, W(u, v)(x) \ne 0, \tag 7$

and likewise,

$W(u, v)(x_0) = 0 \Longrightarrow \forall x, W(u, v)(x) = 0; \tag 8$

these well-known results (6)-(8) establish a bidirectional linkage 'twixt our OP sat091's hypotheses (a) and (b); certainly (b) implies (a), and from what we have done here we have been able to conclude that the weaker (a) yields (b); thus our OP sat091 is correct on both counts.

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