2.36   ODE No. 36

1. Problem in Latex
2. Mathematica input
3. Maple input

$$axy(x{)}^2+y^{\prime }(x)+y(x{)}^3=0$$ ✓ Mathematica : cpu = 0.258517 (sec), leaf count = 195

$$\text{Solve}[\frac{{\text{Ai}}^{\prime }(\frac{\sqrt[3]{-\frac{1}{2}}\sqrt[3]{a}}{y(x)}-\frac{1}{2}\sqrt[3]{-\frac{1}{2}}{a}^{4/3}{x}^2)-{(-\frac{1}{2})}^{2/3}{a}^{2/3}x\text{Ai}(\frac{\sqrt[3]{-\frac{1}{2}}\sqrt[3]{a}}{y(x)}-\frac{1}{2}\sqrt[3]{-\frac{1}{2}}{a}^{4/3}{x}^2)}{{\text{Bi}}^{\prime }(\frac{\sqrt[3]{-\frac{1}{2}}\sqrt[3]{a}}{y(x)}-\frac{1}{2}\sqrt[3]{-\frac{1}{2}}{a}^{4/3}{x}^2)-{(-\frac{1}{2})}^{2/3}{a}^{2/3}x\text{Bi}(\frac{\sqrt[3]{-\frac{1}{2}}\sqrt[3]{a}}{y(x)}-\frac{1}{2}\sqrt[3]{-\frac{1}{2}}{a}^{4/3}{x}^2)}+c_1=0,y(x)]$$ ✓ Maple : cpu = 0.073 (sec), leaf count = 62

$$\{y\left(x\right)=2\frac{a}{a^2{x}^2+2\mathit{R}\mathit{o}\mathit{o}\mathit{t}\mathit{O}\mathit{f}(\mathrm{B}\mathrm{i}\left(\mathit{\_}\mathit{Z}\right)\sqrt[3]{-2a^2}\mathit{\_}\mathit{C}\mathit{1}x+\sqrt[3]{-2a^2}x\mathrm{A}\mathrm{i}\left(\mathit{\_}\mathit{Z}\right)+2{\mathrm{B}\mathrm{i}}^{(1)}\left(\mathit{\_}\mathit{Z}\right)\mathit{\_}\mathit{C}\mathit{1}+2{\mathrm{A}\mathrm{i}}^{(1)}\left(\mathit{\_}\mathit{Z}\right))\sqrt[3]{-2a^2}}\}$$

Hand solution

$$\begin{array}{}\text{(1)}& y^{\prime }\left(x\right)=-ax{y}^2-y^3\end{array}$$

This is Abel first order non-linear. The general form is of Abel first kind is$$y^{\prime }\left(x\right)=f_0\left(x\right)+f_1\left(x\right)y\left(x\right)+f_2\left(x\right)y^2\left(x\right)+f_3\left(x\right)y^3\left(x\right)$$ In this case, $f_0\left(x\right)=0,f_1\left(x\right)=0,f_2\left(x\right)=-ax,f_3\left(x\right)=-1$. 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$$(y+g\left(x\right))y^{\prime }\left(x\right)=f_0\left(x\right)+f_1\left(x\right)y\left(x\right)+f_2\left(x\right)y^2\left(x\right)$$ For $g\left(x\right)\ne 0$.

Looking at (1) again, using the transformation suggested in Kamke $u=\frac{1}{y}-\frac{1}{2}a{x}^2$ or $y=\frac{1}{u+\frac{1}{2}a{x}^2}$Then$$y^{\prime }=\frac{-u^{\prime }-ax}{{(u+\frac{1}{2}a{x}^2)}^2}$$

Equating the above to the RHS of (1) gives$$\begin{array}{rl}\frac{-u^{\prime }-ax}{{(u+\frac{1}{2}a{x}^2)}^2}& =-ax{\left(\frac{1}{u+\frac{1}{2}a{x}^2}\right)}^2-{\left(\frac{1}{u+\frac{1}{2}a{x}^2}\right)}^3\\ -u^{\prime }-ax& =-ax-\frac{1}{u+\frac{1}{2}a{x}^2}\\ \frac{du}{dx}& =\frac{1}{u+\frac{1}{2}a{x}^2}\end{array}$$

Writing as$$\begin{array}{}\text{(2)}& \frac{dx}{du}=u+\frac{1}{2}a{x}^2\end{array}$$ This can now be viewed as reverse Riccati in $x$. Using the standard transformation$$\begin{array}{}\text{(3)}& x=-\frac{z^{\prime }}{z\left(\frac{1}{2}a\right)}=-\frac{2z^{\prime }}{az}\end{array}$$ Hence$$\frac{dx}{du}=-\frac{2}{a}(\frac{z^{\prime \prime }}{z}-\frac{{\left(z^{\prime }\right)}^2}{z^2})$$ Equating this to RHS of (2) gives a second order Airy ODE where the dependent variable is $z$ and the independent variable is $u$$$\begin{array}{rl}-\frac{2}{a}(\frac{z^{\prime \prime }}{z}-\frac{{\left(z^{\prime }\right)}^2}{z^2})& =u+\frac{1}{2}a{(-\frac{2z^{\prime }}{az})}^2\\ -\frac{2}{a}\frac{z^{\prime \prime }}{z}+\frac{2}{a}\frac{{\left(z^{\prime }\right)}^2}{z^2}& =u+\frac{1}{2}a\frac{4{\left(z^{\prime }\right)}^2}{a^2{z}^2}\\ -\frac{2}{a}\frac{z^{\prime \prime }}{z}+\frac{2}{a}\frac{{\left(z^{\prime }\right)}^2}{z^2}& =u+\frac{2}{a}\frac{{\left(z^{\prime }\right)}^2}{z^2}\\ -\frac{2}{a}\frac{z^{\prime \prime }}{z}& =u\\ z^{\prime \prime }\left(u\right)+\frac{a}{2}uz\left(u\right)& =0\end{array}$$

This is Airy ODE whose solution is found using power series method. The solution is$$\begin{array}{}\text{(4)}& z\left(u\right)=C_1\mathrm{AiryAI}(-\frac{1}{2}{2}^{\frac{2}{3}}{a}^{\frac{1}{3}}u)+C_2\mathrm{AiryBI}(-\frac{1}{2}{2}^{\frac{2}{3}}{a}^{\frac{1}{3}}u)\end{array}$$ We now go back to (3) and find $x$$$x=-\frac{2z^{\prime }}{az}$$ Since $$\begin{array}{rl}\frac{d}{du}\mathrm{AiryAI}(-\frac{1}{2}{2}^{\frac{2}{3}}{a}^{\frac{1}{3}}u)& =-\frac{1}{2}{2}^{\frac{2}{3}}{a}^{\frac{1}{3}}\mathrm{AiryAI}(1,-\frac{1}{2}{2}^{\frac{2}{3}}{a}^{\frac{1}{3}}u)\\ \frac{d}{du}\mathrm{AiryBI}(-\frac{1}{2}{2}^{\frac{2}{3}}{a}^{\frac{1}{3}}u)& =-\frac{1}{2}{2}^{\frac{2}{3}}{a}^{\frac{1}{3}}\mathrm{AiryBI}(1,-\frac{1}{2}{2}^{\frac{2}{3}}{a}^{\frac{1}{3}}u)\end{array}$$

Then $$x=-\frac{2}{a}\frac{-C_1\frac{1}{2}{2}^{\frac{2}{3}}{a}^{\frac{1}{3}}\mathrm{AiryAI}(1,-\frac{1}{2}{2}^{\frac{2}{3}}{a}^{\frac{1}{3}}u)-C_2\frac{1}{2}{2}^{\frac{2}{3}}{a}^{\frac{1}{3}}\mathrm{AiryBI}(1,-\frac{1}{2}{2}^{\frac{2}{3}}{a}^{\frac{1}{3}}u)}{C_1\mathrm{AiryAI}(-\frac{1}{2}{2}^{\frac{2}{3}}{a}^{\frac{1}{3}}u)+C_2\mathrm{AiryBI}(-\frac{1}{2}{2}^{\frac{2}{3}}{a}^{\frac{1}{3}}u)}$$ Therefore $\frac{dx}{du}$ is now found from above. Once we find $\frac{dx}{du}$ then $\frac{du}{dx}$ is also found. Using $\frac{du}{dx}=\frac{1}{u+\frac{1}{2}a{x}^2}$ now $u\left(x\right)$ is found. Once $u\left(x\right)$ is found then $y\left(x\right)$ is found from the original transformation $y=\frac{1}{u+\frac{1}{2}a{x}^2}$. This is all now just algebra.