# Series solution of xy^{' '}+2y^{'}-xy=0 I get r(r+1)=0, (r+1)(r+2)c_1=0 and c

Series solution of $x{y}^{{}^{″}}+2{y}^{{}^{\prime }}-xy=0$
I get $r\left(r+1\right)=0,\left(r+1\right)\left(r+2\right){c}_{1}=0$ and
${c}_{n+1}=\frac{{c}_{n-1}}{\left(n+1+r\right)\left(n+r+2\right)}$
The first equation gives the indicial roots $r=-1$ and $r=0$. The case for $r=0$ is fine.
For $r=-1$, I don't see how ${c}_{1}=0$ is implied by the second equation. The way I see it, since $1+r=0$ when $r=-1,{c}_{1}$ is not necessarily zero here, but this leads to a different solution to what is in the text book. What am I missing? Why is ${c}_{1}$ forced to be zero?
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4dorts3x
As you already found a basis solution for $r=0$, you need just one other. Setting ${c}_{1}$ in the case $r=-1$ to some non-zero value will just add a multiple of the first, $r=0$, basis solution, so nothing new gets found.
Note that your DE can be written as $\left(xy\right){}^{″}-\left(xy\right)=0$,
so that the solution you find are ${y}_{1}\left(x\right)=\frac{\text{sinh}\left(x\right)}{x}$ and ${y}_{2}\left(x\right)=\frac{\text{cosh}\left(x\right)}{x}$