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[tex]y'' + p(z)y' + q(z)y=0[/tex]

where y (and its derivatives) is a function of z, z ∈ ℂ.

1)My books says this: In points where both p(z) and q(z) are analytic, y(z) is also analytic. But in points where p(z) or q(z) (or both) aren't analytic, y(z) may not be analytic.

Since y(z) is a differentiable complex function in an open set (it must be for the ODE to make sense, right?), it's holomorphic in that set, and so it's analytic. So I don't understand why y(z) may not be analytic when p or q aren't analytic. Why does y(z) being analytic or not even depends on the behavior of these functions?

2)This is a question about an example.

The problem is:

Find the series solutions, about z = 0, of

[tex]y''(z) + y(z) = 0[/tex]

y can be written as a power series (it's analytic), so:

[tex]y(z)=\sum_{n=0}^{\infty}a_{n}z^{n}[/tex]

Substituting the series in the EDO, we obtain the two-term recurrence relation:

[tex]a_{n+2}=-\frac{a_{n}}{(n+2)(n+1)}, n \geq 0[/tex]

So,

[tex]y(z)=1+z-\frac{z^{2}}{2!}+\frac{z^{3}}{3!}-...[/tex]

Which is equal to:

[tex]y(z) = cos(z) + sin(z)[/tex]

The solution to the EDO is:

[tex]y(z) = c_{1}cos(z) + c_{2}sin(z)[/tex]

Why isn't the solution this though?:

[tex]y(z) = cos(z) + sin(z)[/tex]

We stated that y(z) was a power series, and the power series turned out to be cos(z) + sin(z), so why do we take the linear combination of these 2 functions as the solution? Was the initial assumption that y(z) equals a power series wrong?

Thanks

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# Question about 2nd order linear ODEs series solutions

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