Solution to DEs

Solutions to DEs are functions instead of a number

The solution is a function that when subbed in place of the dependent variable makes the right and left side equal

y^{″} (t) + y (t) = 0

$y = \cos (t)$ is one of the solutions to the DE ( $- \cos (t) + \cos (t) = 0$ ), but not the only one

The general solution of this DE is

y (t) = C_{1} \cos (t) + C_{2} \sin (t)

Initial Value Problems

IVPs are differential equations paired with a set of initial conditions

conditions are used to narrow all infinite possible solution into one
gives a solution that can be represented by a single solution curve (solution trajectory)

\begin{aligned} y^{″} (t) + y (t) = 0 \\ y (0) = - 3 \\ y^{'} (0) = 0 \end{aligned}

Solving the DE with the conditions gives

y (t) = - 3 \cos (t) + 0 \sin (t)

A $n^{t h}$ order linear equation

y^{(n)} (t) + f_{n - 1} (t) y^{(n - 1)} (t) \dots + f_{2} (t) y^{''} (t) + f_{1} (t) y^{'} (t) + f_{0} (t) y (t) = g (t)

With $n$ initial conditions

\begin{aligned} y (t_{0}) & = y_{0} \\ y^{'} (t_{0}) & = y_{0}^{'} \\ \dots \\ y^{(n - 1)} (t_{0}) & = y_{0}^{(n - 1)} \end{aligned}

If $f_{0}, f_{1} \dots, f_{n - 1}$ and $g$ are continuous on $t_{0}$ on an open interval $\to$ $\exists$ one unique solution $y = φ (t)$

theorem only proves that the solution exists throughout the interval open $(a, b)$
p. 21+22