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I need some help to locate and classify the singular points of

$$y′′+ 4y = 0$$

then make conclusion on existence of solutions in powers of $x,(x−1),(x+1)$.

so I know that $p(x)=0$ since there is no $y′$ and $q(x)=4$ so does that mean that $x = 0$ is an irregular singular point? Is that the only one?

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    $\begingroup$ If I understood the definition of singular points right, the equation should have no singular points because p(x) and q(x) are everywhere analytic. The general solution to this equation is $y(x) = c_1 sin(2x) + c_2 cos(2x) $ , so I cannot see a connection with powers of x,x-1,x+1. $\endgroup$ – Peter Mar 24 '14 at 12:07
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This is a constant coefficient ODE. The coefficients are everywhere analytic. It follows that the solution centered at any point is entire. In other words, you can write a power series solution centered at $x=a$ and it will have interval of convergence $(-\infty, \infty)$. In particular, $$ y = c_1\cos(x) + c_2 \sin(x) = c_1\cos(x-a+a) + c_2 \sin(x-a+a)$$ thus $$ y = c_1[\cos(a)\cos(x-a) - \sin(a)\sin(x-a)]+ c_2[\sin(a)\cos(x-a)+\cos(a)\sin(x-a)] $$ and it is easy to find the solution centered at $x=a$ via substitution in the known MacClaurin expansions of sine and cosine from the identity above. Alternatively, and equivalently, just solve via the series substitution technique using $\sum_{n=0}^{\infty}c_n(x-a)^n$. Substitution is faster when you can see one.

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