unique solution on interval containing \(t_0\) where \(p(t), q(t), g(t)\) are continuous
for example, \(t(t-4)y'' + 3ty' + 4y = 2 \quad y(3) = 0, \ y'(3) = -1\)
\(p, q, g\) continuous \(t \neq 0, t \neq 4\)
there is one and only one solution on \(0 < t < 4\)
unique solution means there is one and only one linear combination of \(y_1\) and \(y_2\) for a given set of initial conditions
general solution: \(y = c_1 y_1 + c_2 y_2\)
if \(y_1\) and \(y_2\) are solutions, then
let's sub \(y = c_1 y_1 + c_2 y_2\) in to see it is indeed a solution
So, \( y = c_1 y_1 + c_2 y_2 \) is a solution
Solutions \( y_1, y_2 \)
General solution \( y = c_1 y_1 + c_2 y_2 \)
unique solution if \( \begin{bmatrix} y_1(t_0) & y_2(t_0) \\ y'_1(t_0) & y'_2(t_0) \end{bmatrix} \) is invertible
if so,
\[ \begin{vmatrix} y_1(t_0) & y_2(t_0) \\ y'_1(t_0) & y'_2(t_0) \end{vmatrix} \neq 0 \]
as long as \( W[y_1, y_2](t_0) \neq 0 \), we can always uniquely find \( c_1, c_2 \) for any \( y_0, y'_0 \)
The characteristic equation is:
Thus, the fundamental solutions are:
Notice that \( e^{-8t} \neq 0 \) for any \( t \).
\( \rightarrow \) We can find a unique set of \( c_1, c_2 \) for ANY \( y(t_0) = y_0, y'(t_0) = y'_0 \).
Do we need to know \( y_1 \) and \( y_2 \) to find the Wronskian?
Surprisingly, no!
Suppose \( y'' + py' + qy = 0 \) has \( y_1, y_2 \) as fundamental solutions.
Then:
Multiply the 1st equation by \( -y_2 \) and the 2nd by \( y_1 \):
Now add them together:
Note that the terms in parentheses correspond to the Wronskian and its derivative:
we end up with
\[ w' + pw = 0 \]
linear and separable
\[ w = c e^{\int -p dt} \]