Linear Second Order ODE/y'' - 3 y' + 2 y = 0

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

has the general solution:
 * $y = C_1 e^x + C_2 e^{2 x}$

Proof
Consider the functions:


 * $y_1 \left({x}\right) = e^x$
 * $y_2 \left({x}\right) = e^{2 x}$

We have that:

Putting $e^x$ and $e^{2 x}$ into $(1)$ in turn:

Hence it can be seen that:

are particular solutions to $(1)$.

Calculating the Wronskian of $y_1$ and $y_2$:

So the Wronskian of $y_1$ and $y_2$ is never zero.

Thus from Zero Wronskian of Solutions of Homogeneous Linear Second Order ODE iff Linearly Dependent:
 * $y_1$ and $y_2$ are linearly independent everywhere on $\R$.

We have that $(1)$ is a homogeneous linear second order ODE in the form:
 * $y'' + P \left({x}\right) y' + Q \left({x}\right) y = 0$

where $P \left({x}\right) = -3$ and $Q \left({x}\right) = 2$.

So by Constant Real Function is Continuous:
 * $P$ and $Q$ are continuous on $\R$.

Thus from Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution:
 * $(1)$ has the general solution:
 * $y = C_1 e^x + C_2 e^{2 x}$