Derivative of Solution to Constant Coefficient Homogeneous LSOODE is also Solution

Theorem
Let:
 * $(1): \quad y'' + p y' + q y = 0$

be a constant coefficient homogeneous linear second order ODE.

Let $\map y x$ be a particular solution of $(1)$.

Then its derivative $\map {y'} x$ is also a particular solution of $(1)$.

Proof
By Solution of Constant Coefficient Homogeneous LSOODE, $y$ is in one of the following forms:
 * $y = \begin{cases}

C_1 e^{m_1 x} + C_2 e^{m_2 x} & : p^2 > 4 q \\ & \\ C_1 e^{m_1 x} + C_2 x e^{m_2 x} & : p^2 = 4 q \\ & \\ e^{a x} \paren {C_1 \sin b x + C_2 \cos b x} & : p^2 < 4 q \end{cases}$

Let:
 * $y = A e^{m_1 x} + B e^{m_2 x}$

where $A, B \in \R$.

Then:
 * $y' = m_1 A e^{m_1 x} + m_2 B e^{m_2 x}$

which is also in the form:
 * $y = C_1 e^{m_1 x} + C_2 e^{m_2 x}$

and so is also a particular solution of $(1)$.

Let:
 * $y = A e^{m_1 x} + B x e^{m_1 x}$

where $A, B \in \R$.

Then:

which is also in the form:
 * $y = C_1 e^{m_1 x} + C_2 x e^{m_1 x}$

and so is also a particular solution of $(1)$.

Let:
 * $y = e^{a x} \left({A \sin b x + B \cos b x}\right)$

where $A, B \in \R$.

Then:
 * $y' = m_1 A e^{m_1 x} + m_2 B e^{m_2 x}$

which is also in the form:
 * $y = e^{a x} \paren {C_1 \sin b x + C_2 \cos b x}$

and so is also a particular solution of $(1)$.

Hence the result.