<s>[INST] Solve ODE $y'(y'+y)=x(x+1)$ Solve ODE $$y'(y'+y)=x(x+1)$$
I tried to remove $y'^2$ term by differentiate it wrt x and then replace value in hope that it will turn out some exact form but got stuck after
$$2y'y''-yy'+y''=x-x^2+1$$
How i proceed further or my method is wrong ?
Edit:
Exact problem
If $y'-x\neq0$ is a solution of the differential equation $y'(y'+y)=x(x+1)$ then y(x) is given by

*

*$1-x-e^x$


*$1-x-e^{-x}$


*$1+x+e^x$


*$1+x+e^{-x}$
 [/INST] The equation can be rewritten as $(y')^2+yy'-x^2-x=0.$ One can solve $y'$ in terms of $y$ and $x.$ Notice that the above equation is equivalent to $(2y')^2+2y(2y')-4x^2-4x=0=(2y')^2+2y(2y')+y^2-(y^2+4x^2+4x)=(2y'+y)^2-(y^2+4x^2+4x),$ hence $(2y'+y)^2=y^2+4x^2+4x,$ implying $y^2\geq-(4x^2+4x).$ Notice that $4x^2+4x=4x^2+4x+1-1=(2x+1)^2-1,$ so $y^2\geq1-(2x+1)^2.$ This allows us to say that $2y'+y=\pm\sqrt{y^2+(2x+1)^2-1}.$
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