we have the O.D.E ${ (y') }^{ 3 }+y'=x$ using the Paramter $p=y'\quad (dy=pdx)$
we get ${ p }^{ 3 }+p=x$ taking the differential on both side gives
${ 3p }^{ 2 }+1=dx$ hence we get $\frac { dy }{ p } ={ 3p }^{ 2 }+1\quad dy={ 3p }^{ 3 }+p$ by using $dy=pdx$
Ingeration gives $\int { dy } =\int { { 3p }^{ 3 }\quad dp }+ \int { p\quad dp } $
hence we have the solution $y=\frac { { 3p }^{ 4 } }{ 4 } +\frac { { p }^{ 2 } }{ 2 } $ , ${ p }^{ 3 }+p=x$ in a parametric form and im struggling to convert the parametric form into the general solution.
Maybe after the integration we obtain an additional constant, so I assume we have the solution $y=\frac { { 3p }^{ 4 } }{ 4 } +\frac { { p }^{ 2 } }{ 2 }+C$. Solving this equation for $p^2$, we obtain $p^2=\frac 13\left(–1\pm\sqrt{1-12(y+C)} \right)$. Since $p^2$ is non-negative, the only possible case is $p^2=\frac 13\left(–1+\sqrt{1-12(y+C)} \right)$. Then
$$x=p^3+p=(p^2+1)p=\frac 13\left(2+\sqrt{1-12(y+C)} \right)\sqrt{\frac 13\left(–1+\sqrt{1-12(y+C)}\right)}=$$
$$\frac 1{3\sqrt 3}\left(2+\sqrt{1-12(y+C)} \right)\sqrt{\sqrt{1-12(y+C)}–1}.$$
In particular, we have $x\ge\frac{3\sqrt{3}}2$ and $1-12(y+C)\ge 1$, that is $y\le -C$. Taking these inequalities in the account, we can simplify the last equality as follows.
$$3\sqrt 3x-2=\sqrt{1-12(y+C)} \sqrt{\sqrt{1-12(y+C)}–1}$$
$$(3\sqrt 3x-2)^2=(1-12(y+C))\left(\sqrt{1-12(y+C)}–1\right)$$
$$\frac{(3\sqrt 3x-2)^2}{1-12(y+C)}+1=\sqrt{1-12(y+C)}$$
$$\left(\frac{(3\sqrt 3x-2)^2}{1-12(y+C)}+1\right)^2=1-12(y+C)$$
$$\left(\frac{(3\sqrt 3x-2)^2}{1-12(y+C)} \right)^2+2\frac{(3\sqrt 3x-2)^2}{1-12(y+C)} +12(y+C)=0$$
$$ {(3\sqrt 3x-2)^4}+2 (3\sqrt 3x-2)^2(1-12(y+C)) +12(y+C) (1-12(y+C))^2=0$$
Or, substituting $u=(3\sqrt 3x-2)^2\ge 0$ and $v=1-12(y+C)\ge 1$, we obatin
$$(u+v)^2=v^3.$$