\[ -a x-b+y'(x)+y(x)^2=0 \] ✓ Mathematica : cpu = 0.0909047 (sec), leaf count = 79
\[\left \{\left \{y(x)\to -\frac {\sqrt [3]{a} \text {Bi}'\left (\frac {b+a x}{a^{2/3}}\right )+\sqrt [3]{a} c_1 \text {Ai}'\left (\frac {b+a x}{a^{2/3}}\right )}{-\text {Bi}\left (\frac {b+a x}{a^{2/3}}\right )-c_1 \text {Ai}\left (\frac {b+a x}{a^{2/3}}\right )}\right \}\right \}\] ✓ Maple : cpu = 0.194 (sec), leaf count = 79
\[y \relax (x ) = -\frac {i \left (-i a \right )^{\frac {1}{3}} \left (\AiryAi \left (1, -\frac {a x +b}{\left (-i a \right )^{\frac {2}{3}}}\right ) c_{1}+\AiryBi \left (1, -\frac {a x +b}{\left (-i a \right )^{\frac {2}{3}}}\right )\right )}{\AiryAi \left (-\frac {a x +b}{\left (-i a \right )^{\frac {2}{3}}}\right ) c_{1}+\AiryBi \left (-\frac {a x +b}{\left (-i a \right )^{\frac {2}{3}}}\right )}\]
Hand solution
\begin {align} y^{\prime }\relax (x) +y^{2}\relax (x) -ax-b & =0\nonumber \\ y^{\prime }\relax (x) & =b+ax-y^{2}\relax (x) \tag {1} \end {align}
This is Riccati first order non-linear ODE of the form \begin {equation} y^{\prime }\relax (x) =P\relax (x) +Q\relax (x) y+R\left ( x\right ) y^{2}\relax (x) \tag {2} \end {equation} where in this case \(Q\relax (x) =0,R\relax (x) =-1,P\left ( x\right ) =b+ax\). We can solve this in two ways. If we know one particular solution \(y_{p}\relax (x) \) for (1) then we use the substitution \(y=y_{p}+\frac {1}{u}\) and convert (1) to new associated linear ODE of the form \(u^{\prime }+\left (Q\relax (x) +2R\relax (x) \right ) y_{p}+R\relax (x) =0\). If we do not know a particular solution, then we use the standard substitution \(y=\frac {-u^{\prime }}{uR\relax (x) }=\frac {u^{\prime }}{u}\) since \(R\relax (x) =-1\) and this is what we will do here.
Since \(u^{\prime }=yu\) then \begin {align*} u^{\prime \prime } & =yu^{\prime }+y^{\prime }u\\ & =y\left (yu\right ) +\left (b+ax-y^{2}\right ) u\\ & =y^{2}u+\left (b+ax\right ) u-y^{2}u\\ & =\left (b+ax\right ) u \end {align*}
So we have new second order ODE \begin {equation} u^{\prime \prime }-\left (b+ax\right ) u=0 \tag {3} \end {equation} which we solve for \(u\). This ODE is of the form \(u^{\prime \prime }-q\left ( x\right ) u=0\) which has solutions in terms of Airy function of first \(Ai\relax (x) \) and second kind \(Bi\relax (x) \), where
\begin {align*} Ai\relax (x) & =\frac {1}{\pi }\int _{0}^{\infty }\cos \left (\frac {t^{3}}{3}+xt\right ) dt\\ Bi\relax (x) & =\frac {1}{\pi }\int _{0}^{\infty }\exp \left ( -\frac {t^{3}}{3}+xt\right ) +\sin \left (\frac {t^{3}}{3}+xt\right ) dt \end {align*}
Therefore the solution to (3) is
\[ u\relax (x) =c_{1}Ai\left (\frac {b+ax}{a^{\frac {2}{3}}}\right ) +c_{2}Bi\left (\frac {b+ax}{a^{\frac {2}{3}}}\right ) \]
We need to find \(u^{\prime }\relax (x) \) now. Using \(Ai^{\prime }\left ( x\right ) ,Bi^{\prime }\relax (x) \) for derivative of Airy functions of first and second kind, then
\[ u^{\prime }\relax (x) =c_{1}Ai^{\prime }\left (\frac {b+ax}{a^{\frac {2}{3}}}\right ) a^{\frac {1}{3}}+c_{2}Bi^{\prime }\left (\frac {b+ax}{a^{\frac {2}{3}}}\right ) a^{\frac {1}{3}}\]
Therefore since \(u^{\prime }=yu\) then
\begin {align*} y & =\frac {u^{\prime }}{u}\\ & =\frac {c_{1}A^{\prime }i\left (\frac {b+ax}{a^{\frac {2}{3}}}\right ) a^{\frac {1}{3}}+c_{2}Bi^{\prime }\left (\frac {b+ax}{a^{\frac {2}{3}}}\right ) a^{\frac {1}{3}}}{c_{1}Ai\left (\frac {b+ax}{a^{\frac {2}{3}}}\right ) +c_{2}Bi\left (\frac {b+ax}{a^{\frac {2}{3}}}\right ) } \end {align*}
Let \(C_{1}=\frac {c_{1}}{c_{2}}\) then the above can be written as
\[ y=\frac {C_{1}A^{\prime }i\left (\frac {b+ax}{a^{\frac {2}{3}}}\right ) a^{\frac {1}{3}}+Bi^{\prime }\left (\frac {b+ax}{a^{\frac {2}{3}}}\right ) a^{\frac {1}{3}}}{C_{1}Ai\left (\frac {b+ax}{a^{\frac {2}{3}}}\right ) +Bi\left (\frac {b+ax}{a^{\frac {2}{3}}}\right ) }\]
Reference: Airy function