Find $y(2) $ given $y(x)$ given a separable differential equation Find what $y(2)$ equals if $y$ is a function of $x$ which satisfies:
$x y^5\cdot y'=1$ given $y=6$ when $x=1$
I got $y(2)=\sqrt{6\ln(2)-46656}$
but this answer is wrong can anyone help me figure out the right answer and how I went wrong?

This equation is again separable (you would benefit from learning how to solve separable equations as they are the most simple class of differential equations to solve).
$$
\frac{dy}{dx}xy^5=1\Rightarrow y^5dy=x^{-1}dx\Rightarrow \frac{y^6}{6}=\ln(x)+c
$$
Using boundary condition $y(1)=6$ yields
$$ 
\frac{y^6}{6}=\ln(x)+c\Rightarrow \frac{6^6}{6}=\ln(1)+c\Rightarrow 6^5=c
$$
Then we have solution
$$
\frac{y^6}{6}=\ln(x)+6^5
$$
Now you can plug in whatever value you want. 
