\[ y(x) (a x+b)+y''(x)=0 \] ✓ Mathematica : cpu = 0.0202943 (sec), leaf count = 42
\[\left \{\left \{y(x)\to c_1 \text {Ai}\left (-\frac {b+a x}{(-a)^{2/3}}\right )+c_2 \text {Bi}\left (-\frac {b+a x}{(-a)^{2/3}}\right )\right \}\right \}\]
✓ Maple : cpu = 0.039 (sec), leaf count = 31
\[ \left \{ y \left ( x \right ) ={\it \_C1}\,{{\rm Ai}\left (-{(ax+b){a}^{-{\frac {2}{3}}}}\right )}+{\it \_C2}\,{{\rm Bi}\left (-{(ax+b){a}^{-{\frac {2}{3}}}}\right )} \right \} \]
\begin {equation} y^{\prime \prime }+\left ( ax+b\right ) y=0 \tag {1} \end {equation}
For \(a\neq 0.\) Let \(y=\eta \left ( \xi \right ) ,\xi =ax+b\), hence
\begin {align*} \frac {dy}{dx} & =\frac {d\eta }{d\xi }\frac {d\xi }{dx}\\ & =\frac {d\eta }{d\xi }a \end {align*}
And
\begin {align*} \frac {d^{2}y}{dx^{2}} & =\frac {d}{dx}\left ( \frac {d\eta }{d\xi }a\right ) \\ & =a\frac {d}{dx}\left ( \frac {d\eta }{d\xi }\right ) \\ & =a\left ( \frac {d^{2}\eta }{d\xi ^{2}}\frac {d\xi }{dx}\right ) \\ & =a^{2}\frac {d^{2}\eta }{d\xi ^{2}} \end {align*}
Therefore (1) becomes
\begin {align} a^{2}\frac {d^{2}\eta }{d\xi ^{2}}+\xi \eta \left ( \xi \right ) & =0\nonumber \\ a^{2}\eta ^{\prime \prime }+\xi \eta & =0 \tag {2} \end {align}
This is Airy ODE but with plus sign instead of the normal Airy \(\eta ^{\prime \prime }-\xi \eta =0\). Let
\begin {align*} \eta & =\sum _{n=0}^{\infty }c_{n}\xi ^{n}\\ \eta ^{\prime } & =\sum _{n=0}^{\infty }nc_{n}\xi ^{n-1}=\sum _{n=1}^{\infty }nc_{n}\xi ^{n-1}=\sum _{n=0}^{\infty }\left ( n+1\right ) c_{n+1}\xi ^{n}\\ \eta ^{\prime \prime } & =\sum _{n=0}^{\infty }n\left ( n+1\right ) c_{n+1}\xi ^{n-1}=\sum _{n=1}^{\infty }n\left ( n+1\right ) c_{n+1}\xi ^{n-1}=\sum _{n=0}^{\infty }\left ( n+1\right ) \left ( n+2\right ) c_{n+2}\xi ^{n} \end {align*}
Hence (2) becomes
\begin {align*} a^{2}\sum _{n=0}^{\infty }\left ( n+1\right ) \left ( n+2\right ) c_{n+2}\xi ^{n}+\xi \sum _{n=0}^{\infty }c_{n}\xi ^{n} & =0\\ \sum _{n=0}^{\infty }a^{2}\left ( n+1\right ) \left ( n+2\right ) c_{n+2}\xi ^{n}+\sum _{n=0}^{\infty }c_{n}\xi ^{n+1} & =0\\ \sum _{n=0}^{\infty }a^{2}\left ( n+1\right ) \left ( n+2\right ) c_{n+2}\xi ^{n}+\sum _{n=1}^{\infty }c_{n-1}\xi ^{n} & =0\\ 2a^{2}c_{2}+\sum _{n=1}^{\infty }\left ( a^{2}\left ( n+1\right ) \left ( n+2\right ) c_{n+2}+c_{n-1}\right ) \xi ^{n} & =0 \end {align*}
Hence
\begin {align} 2a^{2}c_{2} & =0\tag {3}\\ a^{2}\left ( n+1\right ) \left ( n+2\right ) c_{n+2}+c_{n-1} & =0\qquad n\geq 1 \tag {4} \end {align}
From (3) and since \(a\neq 0\)
\[ c_{2}=0 \]
From (4)
\[ c_{n+2}=\frac {-c_{n-1}}{a^{2}\left ( n+1\right ) \left ( n+2\right ) }\]
Hence for \(n=3\), we see from the above recurrence equation and because \(c_{2}=0\) that
\[ c_{5}=\frac {-c_{2}}{a^{2}\left ( 4\right ) \left ( 6\right ) }=0 \]
Similarly, for \(n=6\)
\[ c_{8}=\frac {-c_{5}}{a^{2}\left ( 7\right ) \left ( 8\right ) }=0 \]
Similarly, for \(n=9\)
\[ c_{11}=\frac {-c_{8}}{a^{2}\left ( 10\right ) \left ( 11\right ) }=0 \]
And so on. Hence we found so far that for \(n=3,6,9,12,\cdots \) all terms generated which are \(c_{5},c_{8},c_{11},\cdots \) are zero.
Now, for \(n=1\), the recurrence equation gives
\[ c_{3}=\frac {-c_{0}}{a^{2}}\frac {1}{2\cdot 3}\]
For \(n=2\)
\[ c_{4}=\frac {-c_{1}}{a^{2}}\frac {1}{3\cdot 4}\]
For \(n=4\)
\[ c_{6}=\frac {-c_{3}}{a^{2}\left ( 5\right ) \left ( 6\right ) }=c_{0}\left ( \frac {1}{a^{2}}\frac {1}{5\cdot 6}\right ) \left ( \frac {1}{a^{2}}\frac {1}{2\cdot 3}\right ) \]
For \(n=5\)
\[ c_{7}=\frac {-c_{4}}{a^{2}\left ( 6\right ) \left ( 7\right ) }=c_{1}\left ( \frac {1}{a^{2}}\frac {1}{6\cdot 7}\right ) \left ( \frac {1}{a^{2}}\frac {1}{3\cdot 4}\right ) \]
For \(n=7\)
\[ c_{9}=\frac {-c_{6}}{a^{2}\left ( 8\right ) \left ( 9\right ) }=-c_{0}\left ( \frac {1}{a^{2}}\frac {1}{8\cdot 9}\right ) \left ( \frac {1}{a^{2}}\frac {1}{5\cdot 6}\right ) \left ( \frac {1}{a^{2}}\frac {1}{2\cdot 3}\right ) \]
For \(n=8\)
\[ c_{10}=\frac {-c_{7}}{a^{2}\left ( 9\right ) \left ( 10\right ) }=-c_{1}\left ( \frac {1}{a^{2}}\frac {1}{9\cdot 10}\right ) \left ( \frac {1}{a^{2}}\frac {1}{6\cdot 7}\right ) \left ( \frac {1}{a^{2}}\frac {1}{3\cdot 4}\right ) \]
Therefore, in summary, this is what we have so far. For \(n=3,6,9,12,\cdots \) all terms are zero. For \(n=1,4,7,\cdots \) all terms are expressed using \(c_{0}\) and for \(n=2,5,8,\cdots \) all terms are expressed using \(c_{1}\). So there are two arbitrary constants \(c_{0},c_{1}\).
In other words, \(c_{2},c_{5},c_{8},c_{11},\cdots =0\) and \(c_{3},c_{6},c_{9},c_{12},\cdots =f\left ( c_{0}\right ) \) and \(c_{4},c_{7},c_{10},c_{13},\cdots =f\left ( c_{1}\right ) \).
\begin {align*} \eta & =\sum _{n=0}^{\infty }c_{n}\xi ^{n}\\ & =c_{0}+c_{1}\xi ^{1}+c_{2}\xi ^{2}+c_{3}\xi ^{3}+\cdots \\ & =c_{0}+\left ( c_{1}\xi ^{1}\right ) +0-\left ( c_{0}\frac {\xi ^{3}}{a^{2}}\frac {1}{2\cdot 3}\right ) -\left ( c_{1}\frac {\xi ^{4}}{a^{2}}\frac {1}{3\cdot 4}\right ) +0+c_{0}\frac {\xi ^{6}}{a^{4}}\frac {1}{2\cdot 3\cdot 5\cdot 6}+c_{1}\frac {\xi ^{7}}{a^{4}}\frac {1}{3\cdot 4\cdot 6\cdot 7}+0\\ & =c_{0}\left ( 1-\frac {\xi ^{3}}{a^{2}}\frac {1}{2\cdot 3}+\frac {\xi ^{6}}{a^{4}}\frac {1}{2\cdot 3\cdot 5\cdot 6}-\cdots \right ) +c_{1}\left ( \xi -\frac {\xi ^{4}}{a^{2}}\frac {1}{3\cdot 4}+\frac {\xi ^{7}}{a^{4}}\frac {1}{3\cdot 4\cdot 6\cdot 7}-\cdots \right ) \\ & =c_{0}\left ( 1-\frac {1}{a^{2}}\frac {\xi ^{3}}{3!}+\frac {1\cdot 4}{a^{4}}\frac {\xi ^{6}}{6!}-\frac {1\cdot 4\cdot 7}{a^{6}}\frac {\xi ^{9}}{9!}+\cdots \right ) +c_{1}\left ( \xi -\frac {\xi ^{4}}{a^{2}}\frac {2}{4!}+\frac {\xi ^{7}}{a^{4}}\frac {2\cdot 5}{7!}-\frac {\xi ^{10}}{a^{6}}\frac {2\cdot 5\cdot 8}{10!}+\cdots \right ) \end {align*}
Hence
\begin {equation} \eta =c_{0}\left ( \sum _{n=0}^{\infty }3^{n}\left ( \frac {1}{3}\right ) _{n}\frac {\left ( -1\right ) ^{n}}{a^{2n}}\frac {\xi ^{3n}}{\left ( 3n\right ) !}\right ) +c_{1}\left ( \sum _{n=0}^{\infty }3^{n}\left ( \frac {2}{3}\right ) _{n}\frac {\left ( -1\right ) ^{n}}{a^{2n}}\frac {\xi ^{3n+1}}{\left ( 3n+1\right ) !}\right ) \tag {5} \end {equation}
Where \begin {align*} 3^{n}\left ( \frac {1}{3}\right ) _{n} & =\left ( 1\right ) \cdot \left ( 4\right ) \cdot \left ( 7\right ) \ldots \left ( 3n-2\right ) \\ 3^{n}\left ( \frac {2}{3}\right ) _{n} & =\left ( 2\right ) \cdot \left ( 5\right ) \cdot \left ( 8\right ) \ldots \left ( 3n-1\right ) \end {align*}
And
\[ \left ( \frac {1}{3}\right ) _{0}=\left ( \frac {2}{3}\right ) _{0}=1 \]
Equation (5) can be simplified more by moving \(\frac {\left ( -1\right ) ^{n}}{a^{2n}}\) into \(\xi \) as follows
\[ \eta =c_{0}\left ( \sum _{n=0}^{\infty }3^{n}\left ( \frac {1}{3}\right ) _{n}\frac {1}{\left ( 3n\right ) !}\left ( \frac {-\xi }{a^{\frac {2}{3}}}\right ) ^{3n}\right ) +ac_{1}\left ( \sum _{n=0}^{\infty }3^{n}\left ( \frac {2}{3}\right ) _{n}\frac {1}{\left ( 3n+1\right ) !}\left ( \frac {-\xi }{a^{\frac {2}{3}}}\right ) ^{3n+1}\right ) \]
Let \(\left ( \frac {-\xi }{a^{\frac {2}{3}}}\right ) =z\) then the above is
\[ \eta =c_{0}\left ( \sum _{n=0}^{\infty }3^{n}\left ( \frac {1}{3}\right ) _{n}\frac {z^{3n}}{\left ( 3n\right ) !}\right ) +ac_{1}\left ( \sum _{n=0}^{\infty }3^{n}\left ( \frac {2}{3}\right ) _{n}\frac {z^{3n+1}}{\left ( 3n+1\right ) !}\right ) \]
Let \begin {align*} f\left ( \xi \right ) & =\sum _{n=0}^{\infty }3^{n}\left ( \frac {1}{3}\right ) _{n}\frac {\xi ^{3n}}{\left ( 3n\right ) !}\\ g\left ( \xi \right ) & =\sum _{n=0}^{\infty }3^{n}\left ( \frac {2}{3}\right ) _{n}\frac {\xi ^{3n+1}}{\left ( 3n+1\right ) !} \end {align*}
Now looking at definition of \(\operatorname {AiryAI}\left ( z\right ) \) we see
\begin {align*} \operatorname {AiryAI}\left ( z\right ) & =r_{1}f\left ( z\right ) -r_{2}g\left ( z\right ) \\ \operatorname {AiryBI}\left ( z\right ) & =\sqrt {3}\left ( r_{1}f\left ( z\right ) +r_{2}g\left ( z\right ) \right ) \end {align*}
These are Airy functions \(\operatorname {AiryAI}\) and \(\operatorname {AiryBI}\) with appropriate choice of \(c_{0},c_{1}\). See definition of these special functions. Converting back to \(x\) using \(\xi =ax+b\) should result in solution given by CAS. Need to write these final details to make sure. Will finish later.