I've been given a 2nd ODE in the form(adsbygoogle = window.adsbygoogle || []).push({});

y'' + p(x)y' + q(x)y = 0

The equation does not satisfy the test for a unique solution at x_0 = 0, because p and q are not continuous at x_0 (both p and q have x in the denominator, so a value of 0 makes the function discontinuous).

I've two questions.

1) If the uniqueness-existence theorem is not satisfied, can I conclude that the theorem does not guarantee a unique solution, but failure of the test does not exclude there being a unique solution? I.e. a unique solution is possible, but its existence cannot be predicted using the theorem.

2) I've found a general solution to the 2nd ODE of

y_h = c_1*f(x) + c_2*g(x)

I'm given initial conditions of y(0) =0 and y'(0) = 1.

using these values, I get c_2 = 4c_1 - 4.

The other equation I obtain cancels to 0 = 0

From this, can I conclude that no unique solution exists (for those initial values) to the 2nd ODE, since I'm not able to obtain unique values for c_1 and c_2 ?

Thanks

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# 2nd ODE, unique solutions

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