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I am solving this eq: $$w''(z)-\frac{2}{z}w'(z)+w(z)=0$$ using the Frobenius methodology. I found two roots $$w_1=\sum_{k=1}^{\infty}\frac{(-1)^{k}}{4^kk!(-\frac{1}{2})_k} z^{2k} ,w_2=\sum_{k=1}^{\infty}\frac{(-1)^{k}}{4^kk!(\frac{5}{2})_k}z^{2k+3} $$ where $(a)_k$ is the Pochhammer symbol.

Now i want to proove that these two are linearly independent. However when i take their Wronskian it feels impossible to do these calculations. Am i missing something or maybe my solutions are miscalculated??

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    How did you find thses.? – 2017-02-13
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    Like i said with the frobenius method i substut: $w=\sum_{k=0}^{\infty}C_kz^k.$ – 2017-02-13
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    and the second with.? – 2017-02-13
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    the indicial equation has two solutions for λ. – 2017-02-13
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    i forgot up there the power of z is not k it is λ+k – 2017-02-13
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    the indicial equation ensures that two answers are linearly independent. But I have not found a reference yet. – 2017-02-13

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