Suppose that the conditional mean of $Y$ given $x$, that is, $E(Y|x)$, is linear in $x$. Now i'm trying to understand why that means the following, given that the continuous random variable $Y$ follows a normal distribution for each $x$:
$$E(Y|x)=\mu_Y+\rho \dfrac{\sigma_Y}{\sigma_X}(x-\mu_X)$$
I encountered this formula on the top of this page Conditional Distribution of Y Given X. That's a course from Penn. It doesn't define $\rho$, and though it refernces to the previous lessons, notes of the previous lesson don't contain the formula.
From your linearity condition, we have
\begin{equation} \mathbf{E}[Y \, | \, x ] = \alpha + \beta x,\label{eq:lin}\tag{1} \end{equation}
for some constants $\alpha, \beta \in \mathbb{R}$. Letting $f_X(x)$ denote the probability density function of $X$ we have
\begin{align*} \mathbf{E} \big[ \mathbf{E}[Y \, | x ] \big] & = \int f_X(x) (\alpha + \beta x) d x \\ & = \alpha + \beta \int x\, f_X(x) dx \\ & = \alpha + \beta \mu_x \label{eq:ee}\tag{2}, \end{align*}
where we denoted $\mathbf{E}[X] = \mu_X$ as you do in the question. However, by the law of total expectation
$$ \mathbf{E} \big[ \mathbf{E}[Y \, | x ] \big] = \mathbf{E}[Y] = \mu_Y \label{eq:totalex}\tag{3} $$ and hence combining $\ref{eq:ee}$ and $\ref{eq:totalex}$ we have $$ \alpha = \mu_Y - \beta \mu_X, $$ which is to say $$\mathbf{E}[Y \, | \, x] = \mu_Y + \beta(x - \mu_X).$$
It remains then to show that $\beta$ can be expressed in terms of the standard deviation, and correlation of $X$ and $Y$. To do this, you will want to use a similar method, and the law of total variance.