How to solve $(y''')^3-xy'''+y''=0$ I have no idea how to solve this example:
$$(y''')^3-x\cdot y'''+y''=0$$
This ODE is third order, but, what to do with the power: $^3$???
Thanks in advance!
 A: An obvious (and helpful) substitution would be to substitute $v=y''$ to reduce the ODE to first-order:
$$(v')^3-xv'+v=0$$
Notice that you can put it in the form of a Clairaut Equation if you rearrange it:
$$v=xv'-(v')^3$$
Alternatively, if you prefer Leibniz notation:
$$v(x)=x\cdot \frac{dv}{dx}-\left(\frac{dv}{dx}\right)^3$$

The next step would thus be to differentiate both sides wrt. $x$:
$$\frac{dv}{dx}=x\cdot \frac{d^2 v}{dx^2}+\frac{dv}{dx}-3\cdot \left(\frac{dv}{dx}\right)^2\cdot \frac{d^2 v}{dx^2}$$
If you factor this, you should get:
$$\frac{d^2 v}{dx^2}\left(x-3\left(\frac{dv}{dx}\right)^2\right)=0$$
Therefore, you should solve for the following separately:
$$\begin{cases} \frac{d^2v}{dx^2}=0 \\ x-3\cdot \left(\frac{dv}{dx}\right)^2=0 \end{cases}$$
Can you continue?
A: EDIT: notice that this is not THE general solution (as stated before) but it is general within the class of analytic functions. See section 2. for another class.


*

*As proposed in a comment we can try the ansatz


$$y(x)=a x^3+b x^2+c x+d$$
This gives us the 
Then the ODE gives just one relation between the coefficients, viz.
$$b\to -108 a^3$$
so that
$$y(x) =a x^3-108 a^3 x^2+c x+d$$
is a solution. 
And it is even the general solution as in contains three arbitrary constants $a$, $c$, and $d$. These can be used to adapt the solution to initial conditions thus: $$y(0)=d, y'(0)=c, y''(0)=-216 a^3$$


*The power Ansatz


$$y(x) = r x^w$$
leads to
$$w \to \frac{7}{2}, r\to \pm \frac{8}{105 \sqrt{3}} $$
We observe that an integer power in $\sqrt{x}$ appears.
Notice that you can always add a general linear polynomial to $y(x)$
