Given a linear homogenous second-order ODE
$$f_2(x)\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + f_1(x)\frac{\mathrm{d}y}{\mathrm{d}x} + f_0(x)y = 0$$
with initial conditions
$$y(x_0)=0$$ $$\frac{\mathrm{d}y}{\mathrm{d}x}(x_0) = c$$
for some $x_0$, what is the effect of changing $c$?
The background for this is that I have been told that all solutions may be obtained from the solution for $c=1$. (In a context where multiplying the solution by a constant wouldn't make any difference, in case that is relevant.)
I can see that we would want to exclude $c=0$, because this would lead to the trivial solution $\forall x, y(x)=0$, but what does it mean to say that all solutions may be obtained with $y'(x_0)=1$?
Let $y$ be a solution of the differential equation $$f_2(x)\frac{d^2 y}{dx^2}+f_1(x)\frac{dy}{dx}+f_0(x)y=0,$$ satisfying the initial conditions $y(x_0)=0$ and $\frac{dy}{dx}(x_0)=c\ne 0$.
Let $z=\frac{y}{c}$. Then $\frac{dz}{dx}=\frac{1}{c}\frac{dy}{dx}$, and $\frac{dz^2}{dx^2}=\frac{1}{c}\frac{dy^2}{dx^2}$, and therefore $f_2(x)\frac{d^2 z}{dx^2}+f_1(x)\frac{dz}{dx} +f_0(x)z=0$, and $z(x_0)=0$, $\frac{dz}{dx}(x_0)=1$.
Conversely, if
$$f_2(x)\frac{d^2 z}{dx^2}+f_1(x)\frac{dz}{dx} +f_0(x)z=0,$$ with $z(x_0)=0$ and $\frac{dz}{dx}(x_0)=1$, then $y=cz$ is a solution of your original equation, with $y(x_0)=0$ and $\frac{dy}{dx}(x_0)=c$.
The "conversely" part says that all solutions with $c\ne 0$ can be obtained by scaling a solution which has $c=1$.