So I am trying to prove that a linear second order homogeneous differential equation has a solution of the form
$$c_1 e^{m_1 x} + c_2e^{m_2 x}$$
(for real and distinct roots of the auxillary equation, where $m_1$ and $m_2$ are the roots).
1)So I begin with constructing a vector space
$$X=\{f(x): \, f(x) \textrm{ is a solution to the differential equation} \}$$
Then I need to show that this vector is of $2$ dimensions. I can't seem to figure out how to prove that the vector space has two dimensions.
2)Once that is done, we can show that $e^{m_1 x}$ and $e^{m_2 x}$ are spanning vectors with linear independence. Also since we should have by now proved that dimensions of the space is $2$, they form a complete basis set.
Hence, any vector in the space can be represented as a linear combination of the two basis vectors. Which proves that a linear 2nd order homogeneous differential equation has a solution of the form $c_1 e^{m_1 x} + c_2e^{m_2 x}$, with $c_1$ and $c_2$ are arbitrary constants.
Best Answer
We must consider the equation
$y’’+a_1y’+a_2y=0$
with initial conditions $y’(t_0)=m_1$ and $y(t_0)=m_2$.
You can observe that by Cauchy theorem there exists a unique solution of your equation in a neighborhoods of $t_0$. By some theorems of extension of solutions you have that you can extend your unique solution to all $\mathbb{R}$. In this way you get that it is possibile define a bijective map
$\Psi: \mathbb{R}^2\to X$
such that for each couple $(m_1,m_2)$ we consider the solution of the differential equation $\Psi(m_1,m_2)$ of that initial data. You can observe that the map $\Psi$ is linear because the equation is linear. So the map $\Psi$ is an isomorphism and $X$ has dimension 2.
If you want, you can generalize the result to a generic omogeneus equation of degree $n$ using the same argument.