How to prove that the general solution of a homogeneous linear first order ODE has dimension 1? I would like to prove it by using Linear Algebra.
I first defined a Linear Application on an open interval $I$
$$L(.) : C^1(I) \to C(I).$$
$$y(x) \to  y(x)'+p(x)y(x)$$
For a Homogeneous Equation $L(y)=0$, The solution is the Kernel of $L(.)$ which is by definition a Subspace but how can I prove that the dimension is 1?
 A: If $P$ is an antiderivative of $p$ then $y(x) = C e^{-P(x)}$ will satisfy
$$y'(x) = -C e^{-P(x)} p(x) = -p(x) y(x),$$ and hence solve the homogeneous differential equation. Different choices of antiderivatives correspond to different choices of $C$ because $e^{a+K} = e^K e^a = \text{another constant} \times e^a$ is a scalar multiple of $e^a$.
If $f$ is another solution to the equation then
$$
(y/f)' = (y'f - f'y)/f^2 = (-pyf + pfy)/f^2 = 0,
$$
implying that $y/f$ is constant and hence that $f$ is a scalar multiple of $y$.
This analysis does not fully treat the possibility that $f$ might sometimes be zero, where the computation of $(y/f)'$ would not be sensical. In fact, it might make more sense to do the computation with $(f/y)'$ and achieve the same result because $y$ by definition is never zero as long as $C$ is nonzero. I hope this helps. It can be phrased in the language of linear algebra, but does involve engaging with the form of the differential equation. Most results in ODE are similar.
