Thread: How do I get these Equations aligned?

1. How do I get these Equations aligned?

$\frac{dP}{dr}=\frac{-\Big(P+\rho (P)c^2 \Big)} {2} \frac {d\nu} {dr}\]$\\
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$$\Rightarrow d\nu = \frac {-2dP} {P+\rho (P) c^2}$$\\
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$$\nu (r) = - \int_{0}^{P(r)} \frac {2dP}{P+\rho (P) c^2} + \nu (r_b)$$
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$$e^{\nu(r)} &= e^{\nu(r_b) - \int_{0}^{P(r)} \frac {2dP}{P+\rho (P) c^2} }$$
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$$e^ {\nu (r)}= e^{\nu(r_b)} exp \Big[-\int_{0}^{P(r)} \frac {2dP}{P+\rho (P) c^2} \Big]$$

I am new to LaTex, so my codes look very childish, anyway someone please help me put these aligned

2. Re: How do I get these Equations aligned?

\begin{align*} &\frac{dP}{dr}=\frac{-\Big(P+\rho (P)c^2 \Big)} {2} \frac {d\nu} {dr} \Rightarrow \\ \\ & d\nu = \frac {-2dP} {P+\rho (P) c^2} \\ \\ &\nu (r) = - \int_{0}^{P(r)} \frac {2dP}{P+\rho (P) c^2} + \nu (r_b) \\ \\ &\large e^{\nu(r)} = e^{\nu(r_b) - \int_{0}^{P(r)} \frac {2dP}{P+\rho (P) c^2} } \\ \\ &\large e^ {\nu (r)}= e^{\nu(r_b)} exp \Big[-\int_{0}^{P(r)} \frac {2dP}{P+\rho (P) c^2} \Big] \end{align*}

This do it for you?

thanks a lot