2.36   ODE No. 36

\[ a x y(x)^2+y'(x)+y(x)^3=0 \]

✓ Mathematica : cpu = 0.319217 (sec), leaf count = 195

DSolve[a*x*y[x]^2 + y[x]^3 + Derivative[1][y][x] == 0,y[x],x]
 

\[\text {Solve}\left [\frac {\text {Ai}'\left (\frac {\sqrt [3]{-\frac {1}{2}} \sqrt [3]{a}}{y(x)}-\frac {1}{2} \sqrt [3]{-\frac {1}{2}} a^{4/3} x^2\right )-\left (-\frac {1}{2}\right )^{2/3} a^{2/3} x \text {Ai}\left (\frac {\sqrt [3]{-\frac {1}{2}} \sqrt [3]{a}}{y(x)}-\frac {1}{2} \sqrt [3]{-\frac {1}{2}} a^{4/3} x^2\right )}{\text {Bi}'\left (\frac {\sqrt [3]{-\frac {1}{2}} \sqrt [3]{a}}{y(x)}-\frac {1}{2} \sqrt [3]{-\frac {1}{2}} a^{4/3} x^2\right )-\left (-\frac {1}{2}\right )^{2/3} a^{2/3} x \text {Bi}\left (\frac {\sqrt [3]{-\frac {1}{2}} \sqrt [3]{a}}{y(x)}-\frac {1}{2} \sqrt [3]{-\frac {1}{2}} a^{4/3} x^2\right )}+c_1=0,y(x)\right ]\]

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dsolve(diff(y(x),x)+y(x)^3+a*x*y(x)^2 = 0,y(x))
 

\[y \left (x \right ) = \frac {2 a}{a^{2} x^{2}+2 \operatorname {RootOf}\left (\left (-2 a^{2}\right )^{{1}/{3}} \operatorname {AiryBi}\left (\textit {\_Z} \right ) c_{1} x +\left (-2 a^{2}\right )^{{1}/{3}} x \operatorname {AiryAi}\left (\textit {\_Z} \right )+2 \operatorname {AiryBi}\left (1, \textit {\_Z}\right ) c_{1} +2 \operatorname {AiryAi}\left (1, \textit {\_Z}\right )\right ) \left (-2 a^{2}\right )^{{1}/{3}}}\]

Hand solution

\begin{equation} y^{\prime }\left ( x\right ) =-axy^{2}-y^{3}\tag {1}\end{equation}

This is Abel first order non-linear. The general form is of Abel first kind is

\[ y^{\prime }\left ( x\right ) =f_{0}\left ( x\right ) +f_{1}\left ( x\right ) y\left ( x\right ) +f_{2}\left ( x\right ) y^{2}\left ( x\right ) +f_{3}\left ( x\right ) y^{3}\left ( x\right ) \]

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

\[ \left ( y+g\left ( x\right ) \right ) y^{\prime }\left ( x\right ) =f_{0}\left ( x\right ) +f_{1}\left ( x\right ) y\left ( x\right ) +f_{2}\left ( x\right ) y^{2}\left ( x\right ) \]

For \(g\left ( x\right ) \neq 0\).

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 }\left ( u\right ) +\frac {a}{2}uz\left ( u\right ) & =0 \end{align*}

This is Airy ODE whose solution is found using power series method. The solution is

\begin{equation} z\left ( 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 ) \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 \(u\left ( x\right ) \) is found. Once \(u\left ( x\right ) \) is found then \(y\left ( x\right ) \) is found from the original transformation \(y=\frac {1}{u+\frac {1}{2}ax^{2}}\). This is all now just algebra.