Derivative of Solution to Constant Coefficient Homogeneous LSOODE is also Solution

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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 $(1)$.

$\Box$


Let:

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

where $A, B \in \R$.

Then:

\(\ds y'\) \(=\) \(\ds m_1 A e^{m_1 x} + B e^{m_1 x} + B m_1 x e^{m_1 x}\)
\(\ds \) \(=\) \(\ds \paren {m_1 A + B} e^{m_1 x} + B m_1 x e^{m_1 x}\)

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 $(1)$.

$\Box$


Let:

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

where $A, B \in \R$.

Then:

$y' = m_1 A e^{m_1 x} + m_2 B e^{m_2 x}$
\(\ds y'\) \(=\) \(\ds a e^{a x} \paren {A \sin b x + B \cos b x} + e^{a x} \paren {A b \cos b x - B b \sin b x}\)
\(\ds \) \(=\) \(\ds e^{a x} \paren {a A \sin b x + a B \cos b x + A b \cos b x - B b \sin b x}\)
\(\ds \) \(=\) \(\ds e^{a x} \paren {\paren {a B + A b} \cos b x + \paren {a A - b B} \sin b x}\)

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 $(1)$.

$\Box$


Hence the result.

$\blacksquare$


Sources

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