$\mathbb{E}[X^2-Y^2]$ and $\sigma(2X+3Y)$ From Given Expected Values For a homework problem, we are given $\mathbb{E}[X]=2$, $\mathbb{E}[Y]=6$, $\sigma^2(X)=9$, and $\sigma^2(Y)=16$. I don't quite understand expected values, co/variance, and deviations. 
As per my textbook, $\sigma^2(Y)= \mathbb{E}[Y^2]-\mathbb{E}[Y]^2$, but I don't know how to calculate $\mathbb{E}[Y^2]$. I know $\mathbb{E}[Y]^2$ is just $6^2=36$, but according to my textbook $\mathbb{E}[Y^2]$ is NOT equivalent to $\mathbb{E}[Y] \times \mathbb{E}[Y]$. It is only the product of expectations when there are 2 independent random variables, and here there is only 1.
Also, even if I was able to calculate $\mathbb{E}[Y^2]$ (and X^2), it doesn't explain what to do when calculating $\mathbb{E}[X^2-Y^2]$. We aren't given the breakdown of the different independent variables, only what their expected values are. Additionally, for $\sigma$, it says it's the square root of the variance, $\sigma^2$. So I assume to compute $\sigma(2X+3Y)$, we need to first compute $\sigma^2(2X+3Y)$. But, I only understand how to compute the variance when there's 1 independent variable, not 2. 
I'm sure this can't be that difficult, and I'm probably severely overcomplicating this. If someone can please help me out I'd greatly appreciate it. Thank you!
 A: Hint: It is straightforward! 
$$\mathbb{E}(Y^2) = \sigma^2[Y] + \mathbb{E}(Y)^2 = 16  + 6^2 = 52$$
Also, expected value is linear. So, $\mathbb{E}(X+Y) = \mathbb{E}(X) + \mathbb{E}(Y)$.
In Addition: 
$$\sigma^2(\sum_{i=1}^nX_i) = \sum_{i=1}^n\sigma^2(X_i) + \sum_{i\neq j}Cov(X_i,X_j)$$
As we know $X_1$ and $X_2$ are independent, so $Cov(X_1,X_2) = 0$. Hence,
$$\sigma^2(X_1 + X_2) = \sigma^2(X_1) + \sigma^2(X_2)$$
Moreover, we know $\sigma^2(aX) = a^2\sigma^2(X)$. So, 
$$\sigma^2(2X + 3Y) = 4\sigma^2(X) + 9 \sigma^2(Y) = 4\times 9 + 9 \times 16 = 180$$
A: First, you have 
$$
E(X^2-Y^2)=E(X^2)-E(Y^2)=\Big(\sigma^2(X)+[E(X)]^2\Big)-\Big(\sigma^2(Y)+[E(Y)]^2\Big).
$$
Next, because you say "there's 1 independent variable, not 2," I'll assume $X$ and $Y$ are independent, else there wouldn't be enough info to compute $\sigma(3X+3Y)$. Under the assumption, however,  you have:
$$
\sigma^2(2X+3Y)=\sigma^2(2X)+\sigma^2(3Y)=4\sigma^2(X)+9\sigma^2(Y).
$$
The first equality above is where the independence between $X$ and $Y$ is used: independence implies zero covariance (the reverse isn't true however but this problem doesn't concern that).
