Second Degree Third Order Non-linear ODE with Dependent Variable y Missing

Question

The ODE I am having trouble with is:$$Solve: (\frac{d^3y}{dx^3})^2 +x\frac{d^3y}{dx^3}-\frac{d^2y}{dx^2}=0$$ The thing is that I'm not supposed to know non-linear ODEs yet. So, I'm supposing this has to be made linear somehow. I just don't understand how I should convert this into a degree-1 ODE. Help?

Answer 1

hint

If we differentiate, we find

$$2y^{(4)}y^{(3)}+xy^{(4)}=0$$ $$=y^{(4)}\Bigl (2y^{(3)}+x\Bigr) $$

Answer 2

substitute $$\frac{d^2y(x)}{dx^2}=u(x)$$ then we get $$u(x)=\left(\frac{du(x)}{dx}\right)^3+x\frac{du(x)}{dx}$$ differentiate both sides with respect to $x$ $$\frac{du(x)}{dx}=x\frac{d^2u(x)}{dx^2}+\frac{du(x)}{dx}+3\left(\frac{du(x)}{dx}\right)^2\frac{d^2u(x)}{dx^2}$$ $$\frac{d^2u(x)}{dx^2}\left(x+3\left(\frac{du(x)}{dx^2}\right)^2\right)=0$$ and we have to solve $$\frac{d^2u(x)}{dx^2}=0$$ or $$x+3\left(\frac{du(x)}{dx}\right)^2=0$$

Answer 3

A simple hint

$$(y''')^2+xy'''-y''=0$$ Substitute $z=y''$ $$(z')^2+xz'-z=0$$ $$(z')^2+xz' +\frac {x^2} 4 -\frac {x^2} 4-z=0$$ $$(z'+\frac x2)^2 -\frac {x^2} 4-z=0$$ $$(z'+\frac x2)^2 -(z+\frac {x^2} 4)=0$$ Susbtitute $w=z+\frac {x^2} 4 \to w'=z'+\frac x2$ $$\boxed {(w')^2 -w=0}$$ It's easy to solve now