The number of solutions to an $n^{th}$ order differential equation.

For an $n$th order differential equation, why are there always $n$ solutions? Why exactly $n$, not $n - 1, n+1$ or infinite many?

Addendum by LePressentiment :

This is motivated by P176 on Strang's Intro to Lin Alg, 4th Ed. An $n$th order differential equation, how many basis functions does it have? I'd guess $n$, because the diffl eqn have $n$ linearly independent solutions (predicated on Julián Aguirre's answer), all of which look to span the nullspace/solution space of the homogeneous ODE.

In general, a differential equation has an infinite number of solutions. The family of solutions of an equation of order $n$ depends on $n$ constants. If the equation is an homogeneous linear equation of order $n$ then there exist $n$ linearly independent solutions $y_1,\dots,y_n$ such that the general solution is $$ y=C_1y_1+\dots+C_ny_n, $$ where $C_1,\dots C_n$ are constants.

Consider the $n$-th order linear homogeneous equation $$ y^{(n)}+a_{n-1}(x)y^{(n-1)}+\dots+a_1(x)y'+a_0(x)y=0, $$ where the $a_i(x)$ are continuous functions on an interval, that without loss of generality we may assume that contains $x=0$. The theory of existence and uniqueness proves that there are solutions $y_1,\dots,y_n$ such that $$ y_1(0)=1,y_1'(0)=0,y_1''(0)=0,\dots,y^{(n-1)}(0)=0\\ y_2(0)=0,y_2'(0)=1,y_0''(0)=0,\dots,y^{(n-1)}(0)=0\\ y_3(0)=0,y_3'(0)=0,y_3''(0)=1,\dots,y^{(n-1)}(0)=0\\ \dots\\ y_n(0)=0,y_n'(0)=0,y_n''(0)=0,\dots,y^{(n-1)}(0)=1 $$ These solutions are linearly independent, and form a basis of the space of solutions. You can check the details in almost any book on ODE's.