Show that the solutions of the homogeneous linear ODE $$\frac{dy}{dx} + p(x)y = 0$$ on an interval I = [a,b] forms a vector subspace W of the real vector space of continuous function on I. What is the dimension of W?
The ode looks simple (if so) but how to go about the second part? And also, what's the significance of "homogeneous" being given?
the significance of being homogeneous is that the origin( zero function witch is $y(x)=0$) is in the space formed by solutions of ODE witch is a property of Vector Spaces.
also the solution of ODE is as follow: $$\frac{dy}{dx} + p(x)y = 0 \Rightarrow \frac{dy}{y}=-p(x)dx \Rightarrow lny=-\int p(x)dx +Constant \Rightarrow y=Ae^{-\int p(x)dx}$$
what I get from Problem is that we must show the solution is a Vector Subspace on Real Numbers which mean Scalars come from $\mathbb R$.
1- ODE is Homogeneous so zero vector is in solution space apparently.
2-if $y_1=A_1e^{-\int p(x)dx}$ and $y_2=A_2e^{-\int p(x)dx}$ and $c$ is a scalar(a real number in this case) then $y_1+cy_2=(A_1+cA_2)e^{-\int p(x)dx}$ is a solution too. so we are dealing with a vector subspace of the space of continuous functions on $\mathbb R$.
Also dimension of the subspace is one because the only member of the set $\{e^{-\int p(x)dx} \} $ spans the whole space.
