\[ y'(x)-(y(x)+x)^2=0 \] ✓ Mathematica : cpu = 0.0107691 (sec), leaf count = 30
\[\left \{\left \{y(x)\to \frac {1}{c_1 e^{2 i x}-\frac {i}{2}}-x-i\right \}\right \}\]
✓ Maple : cpu = 0.074 (sec), leaf count = 16
\[ \left \{ y \left ( x \right ) =-x-\tan \left ( -x+{\it \_C1} \right ) \right \} \]
\begin {align} y^{\prime }-\left ( y+x\right ) ^{2} & =0\nonumber \\ y^{\prime } & =\left ( y+x\right ) ^{2}\tag {1} \end {align}
This is Riccati first order non-linear ODE of the form. Let \(u=y+x\), then \(u^{\prime }=y^{\prime }+1\) and (1) becomes\begin {align*} u^{\prime }-1 & =u^{2}\\ u^{\prime } & =1+u^{2} \end {align*}
This is separable\begin {align*} \frac {du}{dx}\frac {1}{1+u^{2}} & =1\\ \int \frac {du}{1+u^{2}} & =\int dx\\ \tan ^{-1}u & =x+C\\ u & =\tan \left ( x+C\right ) \end {align*}
Since \(u=y+x\) then
\[ y=\tan \left ( x+C\right ) -x \]