- #1
Forhad3097
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Inthis article, the authors present the inhomogeneous equation
$$\ddot{\phi}_2 + \phi_2 + g_2\phi_1^2 + \omega_1\ddot{\phi}_1 = 0,\tag{11}$$
where
$$ \phi_1 = p_1 \cos(\tau + \alpha), \tag{13}$$
The original solution of the inhomogeneous equation is:
$$\phi_2 = p_2\cos(\tau + \alpha) + q_2\sin(\tau + \alpha) + \frac{g_2}{6}p_1^2[\cos(2\tau + 2\alpha) - 3] $$
$$+ \frac{\omega_1}{4}p_1[2\tau\sin(\tau + \alpha) + \cos(\tau + \alpha)]. \tag{14}$$
but I got
\begin{align}
\phi_2 = p_2 \cos(\tau + \alpha) + q_2 \sin(\tau + \alpha) + \frac{g_2p_1^2}{6} [\cos(2(\tau + \alpha)) -3]+ \omega_1p_1cos(\tau+\alpha)
\end{align}
I got the solution by guessing the particular solution **\begin{align}
\phi_2 = p_2 \cos(\tau + \alpha) + q_2 \sin(\tau + \alpha) + C \cos(2(\tau + \alpha)) + D\sin(2(\tau + \alpha)) + E.
\end{align}**
Where is my mistake?
$$\ddot{\phi}_2 + \phi_2 + g_2\phi_1^2 + \omega_1\ddot{\phi}_1 = 0,\tag{11}$$
where
$$ \phi_1 = p_1 \cos(\tau + \alpha), \tag{13}$$
The original solution of the inhomogeneous equation is:
$$\phi_2 = p_2\cos(\tau + \alpha) + q_2\sin(\tau + \alpha) + \frac{g_2}{6}p_1^2[\cos(2\tau + 2\alpha) - 3] $$
$$+ \frac{\omega_1}{4}p_1[2\tau\sin(\tau + \alpha) + \cos(\tau + \alpha)]. \tag{14}$$
but I got
\begin{align}
\phi_2 = p_2 \cos(\tau + \alpha) + q_2 \sin(\tau + \alpha) + \frac{g_2p_1^2}{6} [\cos(2(\tau + \alpha)) -3]+ \omega_1p_1cos(\tau+\alpha)
\end{align}
I got the solution by guessing the particular solution **\begin{align}
\phi_2 = p_2 \cos(\tau + \alpha) + q_2 \sin(\tau + \alpha) + C \cos(2(\tau + \alpha)) + D\sin(2(\tau + \alpha)) + E.
\end{align}**
Where is my mistake?