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

 A: 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 

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

A: Mhenni's answer is correct but I'd like to emphasize the point where you go astray. You've defined $v(x)=y'(x)$, then you you start thinking of $v$ as a function of $y$ instead of $x$. Just stick with $v$ as a function of $x$. Then you've got
\begin{equation}
v'(x) = {d~\over dx} y'(x) = y''(x)
\end{equation}
Your original differential equation is then just
\begin{equation}
x = 11 (v(x))^{10} v'(x)~.
\end{equation}
From here you can follow the logic in Mhenni's answer.
A: To expand upon Mhenni's answer, notice that $yy' = \frac{1}{2}(y^2)'$. Likewise, $y^2y' = \frac{1}{3}(y^3)'$ and so on. You have something of this form, particularly $11y^{10}y'$. We can recognize this as $(y^{11})'$. Can you take it from here?
