\[ -a x-b+y'(x)+y(x)^2=0 \] ✓ Mathematica : cpu = 0.0715606 (sec), leaf count = 69
\[\left \{\left \{y(x)\to \frac {\sqrt [3]{a} \left (c_1 \text {Ai}'\left (\frac {b+a x}{a^{2/3}}\right )+\text {Bi}'\left (\frac {b+a x}{a^{2/3}}\right )\right )}{c_1 \text {Ai}\left (\frac {b+a x}{a^{2/3}}\right )+\text {Bi}\left (\frac {b+a x}{a^{2/3}}\right )}\right \}\right \}\]
✓ Maple : cpu = 1.201 (sec), leaf count = 79
\[ \left \{ y \left ( x \right ) ={-i\sqrt [3]{-ia} \left ( {{\rm Ai}^{(1)}\left (-{(ax+b) \left ( -ia \right ) ^{-{\frac {2}{3}}}}\right )}{\it \_C1}+{{\rm Bi}^{(1)}\left (-{(ax+b) \left ( -ia \right ) ^{-{\frac {2}{3}}}}\right )} \right ) \left ( {{\rm Ai}\left (-{(ax+b) \left ( -ia \right ) ^{-{\frac {2}{3}}}}\right )}{\it \_C1}+{{\rm Bi}\left (-{(ax+b) \left ( -ia \right ) ^{-{\frac {2}{3}}}}\right )} \right ) ^{-1}} \right \} \]
\begin {align} y^{\prime }\left ( x\right ) +y^{2}\left ( x\right ) -ax-b & =0\nonumber \\ y^{\prime }\left ( x\right ) & =b+ax-y^{2}\left ( x\right ) \tag {1} \end {align}
This is Riccati first order non-linear ODE of the form \begin {equation} y^{\prime }\left ( x\right ) =P\left ( x\right ) +Q\left ( x\right ) y+R\left ( x\right ) y^{2}\left ( x\right ) \tag {2} \end {equation} where in this case \(Q\left ( x\right ) =0,R\left ( x\right ) =-1,P\left ( x\right ) =b+ax\). We can solve this in two ways. If we know one particular solution \(y_{p}\left ( x\right ) \) for (1) then we use the substitution \(y=y_{p}+\frac {1}{u}\) and convert (1) to new associated linear ODE of the form \(u^{\prime }+\left ( Q\left ( x\right ) +2R\left ( x\right ) \right ) y_{p}+R\left ( x\right ) =0\). If we do not know a particular solution, then we use the standard substitution \(y=\frac {-u^{\prime }}{uR\left ( x\right ) }=\frac {u^{\prime }}{u}\) since \(R\left ( x\right ) =-1\) 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 \(Ai\left ( x\right ) \) and second kind \(Bi\left ( x\right ) \), where
\begin {align*} Ai\left ( x\right ) & =\frac {1}{\pi }\int _{0}^{\infty }\cos \left ( \frac {t^{3}}{3}+xt\right ) dt\\ Bi\left ( x\right ) & =\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
\[ u\left ( x\right ) =c_{1}Ai\left ( \frac {b+ax}{a^{\frac {2}{3}}}\right ) +c_{2}Bi\left ( \frac {b+ax}{a^{\frac {2}{3}}}\right ) \]
We need to find \(u^{\prime }\left ( x\right ) \) now. Using \(Ai^{\prime }\left ( x\right ) ,Bi^{\prime }\left ( x\right ) \) for derivative of Airy functions of first and second kind, then
\[ u^{\prime }\left ( x\right ) =c_{1}Ai^{\prime }\left ( \frac {b+ax}{a^{\frac {2}{3}}}\right ) a^{\frac {1}{3}}+c_{2}Bi^{\prime }\left ( \frac {b+ax}{a^{\frac {2}{3}}}\right ) a^{\frac {1}{3}}\]
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