ODE/(D^4 - 1) y = sin x

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Theorem

The second order ODE:

$(1): \quad \paren {D^4 - 1} y' = \sin x$

has a general solution:

$y = C_1 e^x + C_2 e^{-x} + C_3 \sin x + C_4 \cos x + \dfrac {x \cos x} 4$


Proof

First we solve the reduced equation of $(1)$:

$(2): \quad \paren {D^4 - 1} y' = 0$

The auxiliary equation of $(1)$ is:

$(3): \quad: m^4 - 1 = 0$

From Complex $4$th Roots of Unity, the roots of $(2)$ are:

$m_1 = 1$
$m_2 = i$
$m_3 = -1$
$m_4 = -i$

So from Solution of Constant Coefficient Linear nth Order ODE, the general solution of $(2)$ is:

$y_g = C_1 e^x + C_2 e^{-x} + C_3 \sin x + C_4 \cos x$


Because $\sin x$ is already a solution of $(2)$, we try:

$y = A x e^{i x}$

with a view to taking the real part in due course.


\(\ds y\) \(=\) \(\ds A x e^{i x}\)
\(\ds \leadsto \ \ \) \(\ds y'\) \(=\) \(\ds A e^{i x} + A i x e^{i x}\)
\(\ds \) \(=\) \(\ds A \paren {1 + i x} e^{i x}\)
\(\ds \leadsto \ \ \) \(\ds y\) \(=\) \(\ds A \paren {1 + i x} i e^{i x} + A i e^{i x}\)
\(\ds \) \(=\) \(\ds A \paren {2 i - x} e^{i x}\)
\(\ds \leadsto \ \ \) \(\ds y^{\paren 3}\) \(=\) \(\ds A \paren {2 i - x} i e^{i x} - A e^{i x}\)
\(\ds \) \(=\) \(\ds A \paren {-3 - i x} e^{i x}\)
\(\ds \leadsto \ \ \) \(\ds y^{\paren 4}\) \(=\) \(\ds A \paren {-3 - i x} i e^{i x} - i e^{i x}\)
\(\ds \) \(=\) \(\ds A \paren {x - 4 i} e^{i x}\)

It follows that

$-4 i A - 1$

and so:

$A = -\dfrac 1 4$


Thus a particular solution to $(1)$ can be given as:

\(\ds y_p\) \(=\) \(\ds \map \Im {\dfrac 1 4 i x e^{i x} }\)
\(\ds \) \(=\) \(\ds \dfrac {x \cos x} 4\)

and the general solution is:

$y = C_1 e^x + C_2 e^{-x} + C_3 \sin x + C_4 \cos x + \dfrac {x \cos x} 4$

$\blacksquare$


Sources

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