Linearly Independent Solutions to 1st Order Systems

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Theorem

The 1st-order homogeneous linear system of differential equations $x' = A \left({t}\right) x$, expressed with the vectors $x', x: \R \to \R^n$ and the matrix function $A: \R \to M_{n \times n} \left({\R}\right)$, has $n$ linearly independent solutions, and if $ \phi_1 , \phi_2, \dots, \phi_n$ are $n$ linearly independent solutions, then $c_1 \phi_1 + c_2 \phi_2 + \cdots + c_n \phi_n$, where $c_i$ are constants, is a general solution.

Proof

Let $v_1, v_2, \dots, v_n$ be linearly independent vectors in $\R^n$, and let $\phi_i$ be solutions to the IVPs $x' = A \left({t}\right) x, \, x \left({t_0}\right) = v_i$ for $i = 1, 2, \dots, n$.

Suppose the solutions are not independent, i.e. $c_1 \phi_1 + c_2 \phi_2 + \cdots + c_n \phi_n = 0$ for some constants $c_i$ not all zero.

Then:

$c_1 \phi_1 \left({t_0}\right) + c_2 \phi_2 \left({t_0}\right) + \cdots c_n \phi_n \left({t_0}\right) = c_1 v_1 + c_2 v_2 + \cdots + c_n v_n = 0$

meaning the vectors $v_i$ are linearly dependent, a contradiction, so the solutions $\phi_i$ must be linearly independent.

By linearity of the system, every vector function of the form $ x = c_1 \phi_1 + \cdots + c_n \phi_n$ is a solution.

Let $z$ be an arbitrary solution of the system.

Since $\phi_i \left({t_0}\right)$ are linearly independent and count $n$ in number, they form a basis for $\R^n$, hence $z \left({t_0}\right)$ must be a linear combination of those solutions, and then by uniqueness of solutions $z$ is a linear combination of the vector functions $\phi_i$.

This proves this is a general solution.

$\blacksquare$