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Consider the ODE $$y'' - y = 0$$ with solutions $y_{(1)} = c_1 \cosh x + c_2 \sinh x$, or, equivalently, $y_{(2)} = c_1 \exp (+x) + c_2 \exp (-x)$.

If we were to solve the above ODE by Frobenius method, i.e. writing the solution as an ansatz $$y = \sum_{n=0}^\infty c_n x^{n + \lambda},$$ we would obtain the indicial equation $$\lambda (\lambda - 1) = 0$$ with solutions $\lambda = 0,1$ which lead to the solution $y_{(1)}$ above.

I do not understand why is the solution $y_{(1)}$ preferred by the Frobenius method instead of $y_{(2)}$ which would have an indicial equation $\lambda^2 = 0$. At what point does the Frobenius method choose a particular set of linearly independent solutions? I would appreciate any insight on the problem.

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Frobenius method is for solving equations around a singular-regular point. The equation $y''-y=0$ has no singular points. If you want to solve it using power series, you know that it will be of thew form $y(x)=\sum_{n=0}^\infty c_n\,x^n$. The selection of solutions comes from the choice of initial values. For instance, $e^x$ comes from $y(0)=y'(0)=1$ (that is, $c_0=c_1=1$), while $\cosh x$ from $y(0)=1$, $y'(0)=0$ (that is, $c_0=1$, $c_1=0$).

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