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?

$$
x \, y^5\,  y' = 1 \Rightarrow \\
\int y^5 \, dy = \int \frac{dx}{x} \Rightarrow \\
\frac{1}{6} y^6 = \ln(x) + C \quad (x > 0)
$$
Inserting $y(1) = 6$ gives
$$
\frac{1}{6} 6^6 = \ln(1) + C \Rightarrow \\
C = 6^5
$$
so we got the solution:
$$
y^6 = 6 \ln(x) + 6^6 \quad (x > 0)
$$
which yields
$$
y(2)^6 = 6 \ln(2) + 6^6 \iff \\
y(2) = \left(6 \ln(2) + 6^6\right)^{1/6} = 6.000089135987335\dotsb
$$
