\[ a x y(x)^2+y'(x)+y(x)^3=0 \] ✓ Mathematica : cpu = 0.324514 (sec), leaf count = 195
\[\text {Solve}\left [\frac {\text {Ai}'\left (\frac {\sqrt [3]{-\frac {1}{2}} \sqrt [3]{a}}{y(x)}-\frac {1}{2} \sqrt [3]{-\frac {1}{2}} a^{4/3} x^2\right )-\left (-\frac {1}{2}\right )^{2/3} a^{2/3} x \text {Ai}\left (\frac {\sqrt [3]{-\frac {1}{2}} \sqrt [3]{a}}{y(x)}-\frac {1}{2} \sqrt [3]{-\frac {1}{2}} a^{4/3} x^2\right )}{\text {Bi}'\left (\frac {\sqrt [3]{-\frac {1}{2}} \sqrt [3]{a}}{y(x)}-\frac {1}{2} \sqrt [3]{-\frac {1}{2}} a^{4/3} x^2\right )-\left (-\frac {1}{2}\right )^{2/3} a^{2/3} x \text {Bi}\left (\frac {\sqrt [3]{-\frac {1}{2}} \sqrt [3]{a}}{y(x)}-\frac {1}{2} \sqrt [3]{-\frac {1}{2}} a^{4/3} x^2\right )}+c_1=0,y(x)\right ]\] ✓ Maple : cpu = 0.085 (sec), leaf count = 62
\[y \relax (x ) = \frac {2 a}{a^{2} x^{2}+2 \RootOf \left (\AiryBi \left (\textit {\_Z} \right ) \left (-2 a^{2}\right )^{\frac {1}{3}} c_{1} x +\left (-2 a^{2}\right )^{\frac {1}{3}} x \AiryAi \left (\textit {\_Z} \right )+2 \AiryBi \left (1, \textit {\_Z}\right ) c_{1}+2 \AiryAi \left (1, \textit {\_Z}\right )\right ) \left (-2 a^{2}\right )^{\frac {1}{3}}}\]
Hand solution
\begin {equation} y^{\prime }\relax (x) =-axy^{2}-y^{3}\tag {1} \end {equation}
This is Abel first order non-linear. The general form is of Abel first kind is\[ y^{\prime }\relax (x) =f_{0}\relax (x) +f_{1}\relax (x) y\relax (x) +f_{2}\relax (x) y^{2}\relax (x) +f_{3}\relax (x) y^{3}\relax (x) \] In this case, \(f_{0}\relax (x) =0,f_{1}\relax (x) =0,f_{2}\relax (x) =-ax,f_{3}\relax (x) =-1\). Note \(\left (\frac {f_{3}}{f_{2}}\right ) ^{\prime }=\left (\frac {1}{ax}\right ) ^{\prime }=-\frac {1}{a}\). While Abel second kind has the form\[ \left (y+g\relax (x) \right ) y^{\prime }\relax (x) =f_{0}\relax (x) +f_{1}\relax (x) y\relax (x) +f_{2}\relax (x) y^{2}\relax (x) \] For \(g\relax (x) \neq 0\).
Looking at (1) again, using the transformation suggested in Kamke \(u=\frac {1}{y}-\frac {1}{2}ax^{2}\) or \(y=\frac {1}{u+\frac {1}{2}ax^{2}}\)Then\[ y^{\prime }=\frac {-u^{\prime }-ax}{\left (u+\frac {1}{2}ax^{2}\right ) ^{2}}\]
Equating the above to the RHS of (1) gives\begin {align*} \frac {-u^{\prime }-ax}{\left (u+\frac {1}{2}ax^{2}\right ) ^{2}} & =-ax\left ( \frac {1}{u+\frac {1}{2}ax^{2}}\right ) ^{2}-\left (\frac {1}{u+\frac {1}{2}ax^{2}}\right ) ^{3}\\ -u^{\prime }-ax & =-ax-\frac {1}{u+\frac {1}{2}ax^{2}}\\ \frac {du}{dx} & =\frac {1}{u+\frac {1}{2}ax^{2}} \end {align*}
Writing as\begin {equation} \frac {dx}{du}=u+\frac {1}{2}ax^{2}\tag {2} \end {equation} This can now be viewed as reverse Riccati in \(x\). Using the standard transformation\begin {equation} x=-\frac {z^{\prime }}{z\left (\frac {1}{2}a\right ) }=-\frac {2z^{\prime }}{az}\tag {3} \end {equation} Hence\[ \frac {dx}{du}=-\frac {2}{a}\left (\frac {z^{\prime \prime }}{z}-\frac {\left ( z^{\prime }\right ) ^{2}}{z^{2}}\right ) \] Equating this to RHS of (2) gives a second order Airy ODE where the dependent variable is \(z\) and the independent variable is \(u\)\begin {align*} -\frac {2}{a}\left (\frac {z^{\prime \prime }}{z}-\frac {\left (z^{\prime }\right ) ^{2}}{z^{2}}\right ) & =u+\frac {1}{2}a\left (-\frac {2z^{\prime }}{az}\right ) ^{2}\\ -\frac {2}{a}\frac {z^{\prime \prime }}{z}+\frac {2}{a}\frac {\left (z^{\prime }\right ) ^{2}}{z^{2}} & =u+\frac {1}{2}a\frac {4\left (z^{\prime }\right ) ^{2}}{a^{2}z^{2}}\\ -\frac {2}{a}\frac {z^{\prime \prime }}{z}+\frac {2}{a}\frac {\left (z^{\prime }\right ) ^{2}}{z^{2}} & =u+\frac {2}{a}\frac {\left (z^{\prime }\right ) ^{2}}{z^{2}}\\ -\frac {2}{a}\frac {z^{\prime \prime }}{z} & =u\\ z^{\prime \prime }\relax (u) +\frac {a}{2}uz\relax (u) & =0 \end {align*}
This is Airy ODE whose solution is found using power series method. The solution is\begin {equation} z\relax (u) =C_{1}\operatorname {AiryAI}\left (-\frac {1}{2}2^{\frac {2}{3}}a^{\frac {1}{3}}u\right ) +C_{2}\operatorname {AiryBI}\left (-\frac {1}{2}2^{\frac {2}{3}}a^{\frac {1}{3}}u\right ) \tag {4} \end {equation} We now go back to (3) and find \(x\)\[ x=-\frac {2z^{\prime }}{az}\] Since \begin {align*} \frac {d}{du}\operatorname {AiryAI}\left (-\frac {1}{2}2^{\frac {2}{3}}a^{\frac {1}{3}}u\right ) & =-\frac {1}{2}2^{\frac {2}{3}}a^{\frac {1}{3}}\operatorname {AiryAI}\left (1,-\frac {1}{2}2^{\frac {2}{3}}a^{\frac {1}{3}}u\right ) \\ \frac {d}{du}\operatorname {AiryBI}\left (-\frac {1}{2}2^{\frac {2}{3}}a^{\frac {1}{3}}u\right ) & =-\frac {1}{2}2^{\frac {2}{3}}a^{\frac {1}{3}}\operatorname {AiryBI}\left (1,-\frac {1}{2}2^{\frac {2}{3}}a^{\frac {1}{3}}u\right ) \end {align*}
Then \[ x=-\frac {2}{a}\frac {-C_{1}\frac {1}{2}2^{\frac {2}{3}}a^{\frac {1}{3}}\operatorname {AiryAI}\left (1,-\frac {1}{2}2^{\frac {2}{3}}a^{\frac {1}{3}}u\right ) -C_{2}\frac {1}{2}2^{\frac {2}{3}}a^{\frac {1}{3}}\operatorname {AiryBI}\left (1,-\frac {1}{2}2^{\frac {2}{3}}a^{\frac {1}{3}}u\right ) }{C_{1}\operatorname {AiryAI}\left (-\frac {1}{2}2^{\frac {2}{3}}a^{\frac {1}{3}}u\right ) +C_{2}\operatorname {AiryBI}\left (-\frac {1}{2}2^{\frac {2}{3}}a^{\frac {1}{3}}u\right ) }\] Therefore \(\frac {dx}{du}\) is now found from above. Once we find \(\frac {dx}{du}\) then \(\frac {du}{dx}\) is also found. Using \(\frac {du}{dx}=\frac {1}{u+\frac {1}{2}ax^{2}}\) now \(u\relax (x) \) is found. Once \(u\left ( x\right ) \) is found then \(y\relax (x) \) is found from the original transformation \(y=\frac {1}{u+\frac {1}{2}ax^{2}}\). This is all now just algebra.