3.4.10.14 Example $y^{\prime }=\frac{1}{3}\frac{2y+y^3-x^2}{x}$

Solve

$$\begin{array}{rl}{y}^{\prime }& =\frac{1}{3}\frac{2y+y^3-x^2}{x}\\ y^{\prime }& =\omega (x,y)\end{array}$$

Using Maple the infinitesimals are

$$\begin{array}{rl}\xi & =\frac{3}{2x^{\frac{1}{3}}}\\ \eta & =\frac{y}{x^{\frac{4}{3}}}\end{array}$$

(Will need to show how to obtain these). Lets solve this using the integration factor method first. The integrating factor is given by

$$\begin{array}{rl}\mu (x,y)& =\frac{1}{\eta -\xi \omega }\\ & =\frac{1}{\frac{y}{x^{\frac{4}{3}}}-\frac{3}{2x^{\frac{1}{3}}}\left(\frac{1}{3}\frac{2y+y^3-x^2}{x}\right)}\\ & =2\frac{x^{\frac{4}{3}}}{x^2-y^3}\end{array}$$

Then the general solution is

$$\begin{array}{rl}\int \mu (x,y)(dy-\omega dx)& =c_1\\ \int 2\frac{x^{\frac{4}{3}}}{x^2-y^3}(dy-\left(\frac{1}{3}\frac{2y+y^3-x^2}{x}\right)dx)& =c_1\\ \int (2\frac{x^{\frac{4}{3}}}{x^2-y^3}dy-\left(2\frac{x^{\frac{4}{3}}}{x^2-y^3}\right)\left(\frac{1}{3}\frac{2y+y^3-x^2}{x}\right)dx)& =c_1\\ \int (2\frac{x^{\frac{4}{3}}}{x^2-y^3}dy-\left(\frac{2}{3}\frac{x^{\frac{1}{3}}}{x^2-y^3}\right)(2y+y^3-x^2)dx)& =c_1\end{array}$$

Hence we need to find $F(x,y)$ s.t. $dF=(2\frac{x^{\frac{4}{3}}}{x^2-y^3}dy-\left(\frac{2}{3}\frac{x^{\frac{1}{3}}}{x^2-y^3}\right)(2y+y^3-x^2)dx)$ which will make the solution $F=c$. Therefore

$$\begin{array}{rl}dF& =\frac{\partial F}{\partial x}dx+\frac{\partial F}{\partial y}dy\\ & =2\frac{x^{\frac{4}{3}}}{x^2-y^3}dy-\left(\frac{2}{3}\frac{x^{\frac{1}{3}}}{x^2-y^3}\right)(2y+y^3-x^2)dx\end{array}$$

Hence

$$\begin{array}{}\text{(1)}& \frac{\partial F}{\partial x}& =-\frac{2}{3}\frac{x^{\frac{1}{3}}(2y+y^3-x^2)}{x^2-y^3}\text{(2)}& \frac{\partial F}{\partial y}& =2\frac{x^{\frac{4}{3}}}{x^2-y^3}\end{array}$$

Integrating (1) gives

$$\begin{array}{rl}F& =(\int -\frac{2}{3}\frac{x^{\frac{1}{3}}(2y+y^3-x^2)}{x^2-y^3}dx)+g\left(y\right)\\ \text{(3)}& & =\frac{1}{2}{x}^{\frac{4}{3}}+\frac{1}{3}\ln (x^{\frac{4}{3}}+x^{\frac{2}{3}}y+y^2)-\frac{2}{3}\sqrt{3}\arctan \left(\frac{1}{3}\frac{(2x^{\frac{2}{3}}+y)\sqrt{3}}{y}\right)-\frac{2}{3}\ln (x^{\frac{2}{3}}-y)+g\left(y\right)\end{array}$$

Where $g\left(y\right)$ acts as the integration constant but $F$ depends on $x,y$ it becomes an arbitrary function. Taking derivative of the above w.r.t. $y$ gives

$$\begin{array}{}\text{(4)}& \frac{\partial F}{\partial y}=2\frac{x^{\frac{4}{3}}}{x^2-y^3}+g^{\prime }\left(y\right)\end{array}$$

Equating (4,2) gives

$$\begin{array}{rl}2\frac{x^{\frac{4}{3}}}{x^2-y^3}& =2\frac{x^{\frac{4}{3}}}{x^2-y^3}+g^{\prime }\left(y\right)\\ 0& =g^{\prime }\left(y\right)\\ g\left(y\right)& =c_1\end{array}$$

Hence (3) becomes

$$F=\frac{1}{2}{x}^{\frac{4}{3}}+\frac{1}{3}\ln (x^{\frac{4}{3}}+x^{\frac{2}{3}}y+y^2)-\frac{2}{3}\sqrt{3}\arctan \left(\frac{1}{3}\frac{(2x^{\frac{2}{3}}+y)\sqrt{3}}{y}\right)-\frac{2}{3}\ln (x^{\frac{2}{3}}-y)+c_1$$

Therefore the solution is

$$\begin{array}{rl}F& =c\\ \frac{1}{2}{x}^{\frac{4}{3}}+\frac{1}{3}\ln (x^{\frac{4}{3}}+x^{\frac{2}{3}}y+y^2)-\frac{2}{3}\sqrt{3}\arctan \left(\frac{1}{3}\frac{(2x^{\frac{2}{3}}+y)\sqrt{3}}{y}\right)-\frac{2}{3}\ln (x^{\frac{2}{3}}-y)& =c_2\end{array}$$

Where constants $c_1,c$ were combined into $c_2$. Now this ode will be solved using direct symmetry by converting to canonical coordinates.  This is done by using the standard characteristic equation by writing

$$\begin{array}{rl}\frac{dx}{\xi }& =\frac{dy}{\eta }=dS\\ \frac{dx}{\frac{3}{2x^{\frac{1}{3}}}}& =\frac{dy}{\frac{y}{x^{\frac{4}{3}}}}=dS\end{array}$$

First pair of ode’s give

$$\frac{dy}{dx}=\frac{\frac{y}{x^{\frac{4}{3}}}}{\frac{3}{2x^{\frac{1}{3}}}}=\frac{2}{3x}y$$

Hence

$$y=c_1{x}^{\frac{2}{3}}$$

Therefore

$$R=y{x}^{-\frac{2}{3}}$$

And

$$dS=\frac{dx}{\xi }=\frac{2}{3}{x}^{\frac{1}{3}}dx$$

Integrating gives

$$\begin{array}{rl}S& =\int \frac{2}{3}{x}^{\frac{1}{3}}dx\\ & =\frac{1}{2}{x}^{\frac{4}{3}}+c_1\\ & =\frac{1}{2}{x}^{\frac{4}{3}}\end{array}$$

By choosing $c_1=0$. Now the ODE $\frac{dS}{dR}=F\left(R\right)$ is found from

$$\begin{array}{rl}\frac{dS}{dR}& =\frac{\frac{dS}{dx}+\omega (x,y)\frac{dS}{dy}}{\frac{dR}{dx}+\omega (x,y)\frac{dR}{dy}}\\ & =\frac{S_x+\omega (x,y)S_y}{R_x+\omega (x,y)R_y}\end{array}$$

But $S_x=\frac{2}{3}{x}^{\frac{1}{3}},R_x=-\frac{2}{3}y{x}^{-\frac{5}{3}},S_y=0,R_y=x^{-\frac{2}{3}}$. Substituting these into the above and simplifying gives

$$\begin{array}{rl}\frac{dS}{dR}& =\frac{\frac{2}{3}{x}^{\frac{1}{3}}}{-\frac{2}{3}y{x}^{-\frac{5}{3}}+\omega (x,y)x^{-\frac{2}{3}}}\\ & =\frac{\frac{2}{3}{x}^{\frac{1}{3}}}{-\frac{2}{3}y{x}^{-\frac{5}{3}}+\left(\frac{1}{3}\frac{2y+y^3-x^2}{x}\right)x^{-\frac{2}{3}}}\\ & =-2\frac{x^2}{x^2-y^3}\end{array}$$

But $R=y{x}^{-\frac{2}{3}}$ or $y=R{x}^{\frac{2}{3}}$. The above becomes

$$\begin{array}{rl}\frac{dS}{dR}& =-2\frac{x^2}{x^2-R^3{x}^2}\\ & =\frac{-2}{1-R^3}\end{array}$$

Which is a quadrature. Solving gives

$$\begin{array}{rl}\int dS& =\int \frac{-2}{1-R^3}dR\\ S& =-\frac{1}{3}\ln (R^2+x+1)-\frac{2}{3}\sqrt{3}\arctan \left(\frac{1}{3}(1+2R)\sqrt{3}\right)+\frac{2}{3}\ln (R-1)+c_1\end{array}$$

Converting back to $x,y$ gives

$$\begin{array}{rl}\frac{1}{2}{x}^{\frac{4}{3}}& =-\frac{1}{3}\ln ({\left(y{x}^{-\frac{2}{3}}\right)}^2+x+1)-\frac{2}{3}\sqrt{3}\arctan \left(\frac{1}{3}(1+2\left(y{x}^{-\frac{2}{3}}\right))\sqrt{3}\right)+\frac{2}{3}\ln (\left(y{x}^{-\frac{2}{3}}\right)-1)+c_1\\ \frac{1}{2}{x}^{\frac{4}{3}}& =-\frac{1}{3}\ln (y^2{x}^{-\frac{4}{3}}+x+1)-\frac{2}{3}\sqrt{3}\arctan \left(\frac{1}{3}(1+2y{x}^{-\frac{2}{3}})\sqrt{3}\right)+\frac{2}{3}\ln (y{x}^{-\frac{2}{3}}-1)+c_1\end{array}$$