This is in reference to the paper : [Paper][1]
Assume now that the theorem is true for some positive integer $n$. Consider a homogeneous linear differential equation of order $n+1$, $$ y^{(n+1)}=b_{n} y^{(n)}+b_{n-1} y^{(n-1)}+\cdots+b_{0} y $$ where each $b_{k}, k=0,1,2, \ldots, n,$ is a function continuous in $R .$ Let $S_{R}^{n+1}$ be the solution set of and let $\operatorname{dim}\left(S_{R}^{n+1}\right)=n+1$
Let $y_{1}, y_{2}, \ldots, y_{n+1}$ be a basis of $S_{R}^{n+1}$. Let $z_{0} \in R$ be such that $y_{1}\left(z_{0}\right) \neq 0$. Let $D$ be a neighborhood of $z_{0}$ such that $y_{1}(z) \neq 0$ in $D$. Then each of the $n$ functions $y_{2} / y_{1}$ $y_{3} / y_{1}, \ldots, y_{n+1} / y_{1}$ is analytic in $D$. Let $Y_{2}, Y_{3}, \ldots, Y_{n+1}$ be $n$ functions analytic in $D$ such that $$ Y_{k}=\left(y_{k} / y_{1}\right)^{\prime}, \quad k=2,3, \ldots, n+1 $$ It is easy to see that $Y_{2}, Y_{3}, \ldots, Y_{n+1}$ are also linearly independent in $D
My question is how do we show that $Y_{2}, Y_{3}, \ldots, Y_{n+1}$ are linearly independent. Do I use the fact that $y_1,y_2.....y_{n+1}$ are linearly independent. I tried for the case of $Y_{2}, Y_{3}$ and $n = 2$ but it seems to be pretty tedious to expand everything. Any hints on how to go about it? [1]: https://www.jstor.org/stable/2320224
Suppose $\sum_{k=2}^{n+1} c_k Y_k = 0$. Then $$ \left(\frac{\sum_{k=2}^{n+1} c_ky_k}{y_1}\right)'=0. $$ Thus, we must have the inside a constant function, say $C$.
Then $$ \frac{\sum_{k=2}^{n+1} c_ky_k}{y_1}=C. $$ This gives a linear relation $$ \sum_{k=2}^{n+1}c_ky_k - Cy_1 = 0. $$ Since $y_1,\ldots, y_{n+1}$ are linearly independent, we must have $$ C=c_2=\cdots=c_{n+1}=0. $$ This shows that $Y_2, \ldots, Y_{n+1}$ are linearly independent.