Solving the differential equation: $ x = 11 (y')^{10} \cdot y'' $ $$ x = 11 (y')^{10} \cdot y'' $$
I proceed as follows:
$$
\begin{align}
\text{Assume } v \equiv v(y) &= \dfrac{dy}{dx} \\
\implies \dfrac{d^2y}{dx^2} &= \dfrac{dv}{dy} \cdot \dfrac{dy}{dx} \\
&= v \cdot \dot{v} \tag{ $ \frac{dv}{dy} = \dot{v} $ } \\
\therefore \qquad x &= 11 v^{10} \dot{v}
\end{align}
$$
Solving which, I get: $ v^{11} = xy + c $. $ c $ being the constant of integration. What to do now?

How do I solve:
$$ \dfrac{dy}{dx} = \left( xy + c \right)^{\frac{1}{11}} $$


You made a mistake in your derivations. Note that

$$ 11 v^{10}v' = x \implies v^{11} = \frac{x^2}{2}+c_1. $$

Now, one possible solution of the last equation is

$$ v  = \left(\frac{x^2}{2}+c_1 \right)^{\frac{1}{11}},$$

which implies 

$$ \frac{dy}{dx} = \left(\frac{x^2}{2}+c_1 \right)^{\frac{1}{11}}. $$

