Differential Equation $y'' + 8y' - 9y = 0$ and Wronskian given that $$y(1) = 1, y'(1) = 0$$
I have solved the general solution for this homogenous equation as follow:
which is $$r^2 + 8r -9  = 0$$
$$ r = -9, 1$$
$$y = C_1e^t + C_2e^{-9t}$$
$$y' = C_1e^t -9C_2e^{-9}$$
Now, finding the coefficients $C_1, C_2$. I am sure using regular substitution method will give the same answer but the textbook seems to use something called "Superposition" (?) using Wronskian determinants (or something along this line, I don't quite understand it)

How does this work? 
 A: If you have a solution of the form:
$y(t) = c_1 y_1(t) + c_2 y_2(t), y(t_0) = y_0, y'(t_0) = y'_0$
We want to find $c_1$ and $c_2$ suct that:
$c_1 y_1(t_0) + c_2y_2(t_0) = y_0$, and
$c_1y'_1(t_0) + c_2y'_2(t_0) = y'_0$
We can write this in matrix form as:
$\begin{bmatrix} y_1(t_0) & y_2(t_0)\\ y'_1(t_0) & y'_2(t_0)\end{bmatrix} \cdot \begin{bmatrix}c_1\\c_2\end{bmatrix} = \begin{bmatrix}y_0\\y'_0\end{bmatrix}$
Since our solutions are purportedly linearly independent, we know we can solve this system.
Now, how do you solve for $c_1$ and $c_2$ for this?
You can take the inverse since these are linearly independent and then pre-multiply that by the RHS. For example, look at the denominator of $c_1$ and $c_2$ and note that that is the determinant of the matrix. You will get the solution as shown in your post above.
Now, give it go with the real numbers and using this method and make sure it gives you same the result for $c_1$ and $c_2$.
Note that this is exactly as you do with the initial conditions, it is only being written more generally using this approach, but they are obviously identical.
Does that make sense?
