Consider a homogeneous 2nd order linear differential equation
$$a(x)y''(x) + b(x)y'(x) + c(x)y(x)=0,$$
where $a$, $b$, and $c$ are given functions of $x$.
Let $V$ be the set of all real solutions $y(x)$ of this equation. Prove that $V$ is a vector space and show that it’s 2-dimensional and isomorphic to $\mathbb{R}^2$.
Best Answer
Hints Certainly the set $\Omega$ of all real-valued functions on $\mathbb{R}$ is a vector space and $V \subseteq \Omega$. So we only need to prove $V$ is a subspace, i.e. that $0 \in V$ and $V$ is closed under addition and constant multiplication. That $0 \in V$ is straight-forward (can you prove it)? Now assume $x,y \in V$ and let $k \in \mathbb{R}$. Can you prove $kx \in V$ and $x+y \in V$?
That $V$ is 2-dimensional will require showing that its basis has 2 elements, i.e. that only 2 things in $V$ can be chosen linearly independent. How do you test for functions to be linearly independent?
Isomorphism to $\mathbb{R}^2$ is easy. Once you prove it's 2-dimensional, pick $x,y \in V$ as its basis and consider the function $f:V \to \mathbb{R}^2$ which maps any $z = c_1 x + c_2 y \in V$ to $(c_1,c_2) \in \mathbb{R}^2$ for any $c_1,c_2 \in \mathbb{R}$...