Linear Second Order ODE/y'' - 3 y' + 2 y = 0/y(0) = -1, y'(0) = 1

Theorem
The second order ODE:
 * $(1): \quad y'' - 3 y' + 2 y = 0$

with initial conditions:
 * $y \left({0}\right) = -1$
 * $y' \left({0}\right) = 1$

has the solution:
 * $y = -3 e^x + 2 e^{2 x}$

Proof
From Second Order ODE: $y'' - 3 y' + 2 y = 0$, the general solution of $(1)$ is:
 * $y = C_1 e^x + C_2 e^{2 x}$

Differentiating $x$:
 * $y' = C_1 e^x + 2 C_2 e^{2 x}$

Thus for the initial conditions:

and:

Hence the result.