For linear ODE we can tell number of solutions:
The linear ODE for $y(x)$ of $n$th order has $n$ linearly independent solutions and as many undetermined constants. An ODE is linear if $y,y',y'',...,y^{(n)}$ occur linearly.
So the ODE $y'+P(x)y=Q(x)$ has only one solution and one undetermined constant. $y''+y=0$ has many pairs of linearly independent solutions: $(\sin x, \cos x), (e^{ix},e^{-ix}),[\sin(x-a), \cos(x-a)]$. In case of many pairs one member of one pair can be written as a linear combination of other two pairs.
Number of constants in a solution decide the oder of the ODE. In your case $y=C e^{Ax}$, will satisfy a 3nd order ODE. To know it you have to eliminate the constants using differentiation.
$$y=C e^{Ax} \implies ye^{-Ax}=C \implies y'e^{-Ax}-Aye^{-Ax}=0 \implies y'/y=A \implies yy''-y'^2=0$$
This second order non-linear ODE, according to me we cannot comment about its number of solutions. The solution of the non-linear ODE $y''=y'^2$ is $y=A-\log(x+B)$: two constants but single solution.
I hope that it helps.