Particular solution of a differential equation I have a question regarding a certain differential equation.
What is the format of the particular solution of the following differential equation:
$$\frac{d^6 y}{dt^6}+y= \cos t$$
Thank you
 A: Since, it's 6$^{th}$ order derivative and function itself, when resulting expression is simple $\cos$ function, you could have guess it as $A(\cos t + \sin t)$, since after differentiation 6 times you'll get linear combination of $\cos$ and $\sin$ again. But, both $\cos t$ and $\sin t$ are already a solutions of homogeneous part, so it would lead you to nowhere. So, technique is the same $y_p = At(\cos t + \sin t) $
$$
\begin{align*}
y_p &= At(\cos t + \sin t) \\
y_p' &= A \left[ (t+1) \cos t - (t-1)\sin t\right] \\
y_p'' &= -A \left[ (t-2)\cos t + (t+2)\sin t\right] \\
y_p''' &= -A \left[ (t+3) \cos t + (t-3)\sin t\right] \\
y_p^{(IV)} &= A \left[ (t-4) \cos t + (t+4)\sin t\right] \\
y_p^{(V)} &= A \left[ (t+5) \cos t - (t-5)\sin t\right] \\
y_p^{(VI)} &= -A \left[ (t-6) \cos t + (t+6)\sin t\right]
\end{align*}
$$
So ODE can be rewritten as
$$
y^{(VI)} + y = 6A(\cos t - \sin t) = \cos t
$$
You might see, that presence of $\sin t$ makes this system to have no solution for $A$. In order to get rid of it, just remove $\cos t$ from your guess. So you'll get
$$
y^{(VI)} + y = 6A\cos t = \cos t \\
A = \frac 16
$$
So $y_p = \frac 16 t\sin t$
