Wronskian to prove linear independence

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Homework Statement


Let [itex]v_1,v_2[/itex] be any two solutions of the differential equation [itex]y''+ay'+by=0[/itex] such that [itex]\frac {v_2}{v_1}[/itex] is not constant, and let f(x) be any solution of the differential equation as well.

Use the properties of the Wronskian to prove that constants [itex]c_1,c_2[/itex] exist such that:

[tex]c_1 v_1(0) + c_2 v_2(0) = f(0), \qquad c_1 v_1 '(0) + c_1 v_1 '(0) = f' (0)[/tex]

Homework Equations


Here are the relevant properties of the Wronskian, defined as [itex]W(x)=v_1(x) v_2 '(x) - v_2(x)v_1 '(x)[/itex]:

Let W be the Wronskian of two solutions [itex]v_1, v_2[/itex] of the differential equation [itex]y'' + ay' +by =0[/itex].
All the following holds:
[tex]W' +aW =0[/tex]
[tex]W(x) = W(0)e^{-ax}[/tex]
[tex]W(0) = 0 \iff \frac{v_2}{v_1} \text{is constant}[/tex]

The Attempt at a Solution



[itex]\frac{v_2}{v_1}[/itex] is not constant, so [itex]W(0) \ne 0[/itex], and therefore for some constant [itex]d[/itex] we have
[tex]dW(0)=f(0)[/tex]
[tex]d(v_1(0) v_2 '(0) - v_2(0)v_1 '(0)) = f(0)[/tex]
[tex][dv_2'(0)]v_1(0) + [-dv_1'(0)]v_2(0) = f(0)[/tex]

So for our solution, [itex]c_1 = dv_2'(0)[/itex] and [itex]c_2 = -dv_1'(0)[/itex], but this leads to

[tex][dv_2'(0)]v_1'(0) + [-dv_1'(0)]v_2'(0) = f'(0)=0[/tex]

Which is not always true.
 
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