Equation formatting\Alinging

Equation formatting\Alinging

According the question I have written the file but I got wrong somewhere.
Can you please check?
\documentclass{article}
\usepackage{amsmath}
\begin{document}
\begin{align}
&\begin{aligned}
& \ddot\phi_3&+\phi_3+2g_2p_1 \cos(\tau+\alpha)[p_2\cos(\tau + \alpha) +
q_2\sin(\tau + \alpha)\\
& + \frac{g_2}{6}p_1^2[\cos(2\tau + 2\alpha) -
3]]+g_3p^3_1\cos^3(\tau+\alpha)-\cos(\tau+\alpha) [-p_1 (\nabla
\alpha)^2+\Delta p_1]\\
&-\sin(\tau+\alpha) [p_1\Delta \alpha+2\nabla \alpha \nabla
p_1]\qquad+\omega_2p_1\cos(\tau+\alpha)=0
\end{aligned}\nonumber\\
&\begin{aligned}
& \ddot\phi_3&+\phi_3+g_2p_1 p_2 [1+\cos2(\tau+\alpha)] + g_2p_1
q_2\sin2(\tau+\alpha) \\
& + \frac{2 g_2^2 p_1^3}{6} \cos(\tau +\alpha)[2\cos^2(\tau+\alpha)-4] \\
& + g_3p_1^3[\frac{1}{4}(3\cos(\tau +\alpha)+\cos3(\tau+\alpha))]\\
& -\cos(\tau+\alpha)\left[\Delta p_1--p_1 (\nabla \alpha)^2\right] \\
& +\sin(\tau+\alpha)\left[p_1\Delta \alpha+2 \nabla p_1 \nabla
\alpha\right]\\
& + \omega_2p_1\cos(\tau+\alpha) =0
\end{aligned}\nonumber\\
&\begin{aligned}
& \ddot\phi_3&+\phi_3+g_2p_1 p_2 + g_2p_1 p_2 \cos2(\tau+\alpha)] +
g_2p_1 q_2\sin2(\tau+\alpha)\\
& + \frac{2 g_2^2 p_1^3}{3} \cos^3(\tau +\alpha)-\frac{8 g_2^2 p_1^3}{6}
\cos(\tau +\alpha) \\
& +g_3p_1^3[\frac{1}{4}(3\cos(\tau +\alpha)+\cos3(\tau+\alpha))]\\
& -\cos(\tau+\alpha)[\Delta p_1--p_1 (\nabla \alpha)^2] \\
& +\sin(\tau+\alpha)\left[p_1\Delta \alpha+2 \nabla p_1 \nabla
\alpha\right]+\omega_2p_1\cos(\tau+\alpha)=0
\end{aligned}\\
&\begin{aligned}
& \ddot\phi_3&+\phi_3+g_2p_1 p_2 + g_2p_1 p_2 \cos2(\tau+\alpha)] +
g_2p_1 q_2\sin2(\tau+\alpha) \\
& + \frac{2 g_2^2 p_1^3}{3}[\frac{1}{4}(3\cos(\tau
+\alpha)+cos3(\tau+\alpha))] -\frac{8 g_2^2 p_1^3}{6} \cos(\tau +\alpha)
\\
& +g_3p_1^3[\frac{1}{4}(3\cos(\tau +\alpha)+\cos3(\tau+\alpha))] \\
& -\cos(\tau+\alpha)[\Delta p_1-p_1 (\nabla \alpha)^2] \\
&+\sin(\tau+\alpha)[p_1\Delta \alpha+2 \nabla p_1 \nabla
\alpha]+\omega_2p_1\cos(\tau+\alpha)=0
\end{aligned} \\
&\begin{aligned}
&\ddot\phi_3&+\phi_3+\sin(\tau+\alpha)[p_1\Delta \alpha+2 \nabla p_1
\nabla \alpha] \\
& -\cos(\tau+ \alpha)\left[\Delta p_1 -p_1 (\nabla
\alpha)^2+\frac{5}{6}g_2^2p_1^3- \frac{3}{4}g_3p_1^3-p_1 +\omega_2
p_1\right] \\
& +\frac{ p_1^3}{12}(2g_2^2+3g_3)\cos3(\tau + \alpha) \\
& +g_2 p_1 [p_2+ p_2 \cos2(\tau +\alpha)+q_2 \sin2(\tau+\alpha)] =0
\end{aligned}\nonumber\\
&\begin{aligned}\label{eq10}
& \ddot\phi_3&+\phi_3+\sin(\tau+\alpha)[p_1\Delta \alpha+2 \nabla p_1
\nabla \alpha] \\
& -\cos(\tau+ \alpha)[\Delta p_1-p_1 (\nabla \alpha)^2 + \lambda
p_1^3-p_1+\omega_2 p_1] \\
& +\frac{p_1^3}{12}(2g_2^2+3g_3)\cos3(\tau + \alpha) \\
& +g_2 p_1 [p_2+ p_2 \cos2(\tau +\alpha)+q_2 \sin2(\tau+\alpha)]=0
\end{aligned}
\end{align}
\end{document}