The definition used here for the general form of a rational ODE is:

rational_ode := diff(y(x),x) = P1(x,y(x))/P2(x,y(x));

$\mathrm{rational\_ode}≔\frac{\ⅆ}{\ⅆx}y\left(x\right)=\frac{\mathrm{P1}\left(x\,y\left(x\right)\right)}{\mathrm{P2}\left(x\,y\left(x\right)\right)}$

where P1 and P2 are arbitrary bivariate polynomials in x and y(x). This type of ODE cannot be solved in a general manner. In some cases, the symmetry method succeeds in finding rational symmetries for the problem (see dsolve,Lie).

$\mathrm{with}\left(\mathrm{DEtools}\,\mathrm{odeadvisor}\,\mathrm{symgen}\right)$

$\left[\mathrm{odeadvisor}\,\mathrm{symgen}\right]$

$\mathrm{ode}\left[1\right]≔\left(2x^3{y\left(x\right)}^3-x\right)\mathrm{diff}\left(y\left(x\right)\,x\right)+2x^3{y\left(x\right)}^3-y\left(x\right)$

${\mathrm{ode}}_1≔\left(2x^3{y\left(x\right)}^3-x\right)\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)+2x^3{y\left(x\right)}^3-y\left(x\right)$

$\mathrm{odeadvisor}\left(\mathrm{ode}\left[1\right]\right)$

$\left[\mathrm{\_rational}\right]$

$\mathrm{symgen}\left(\mathrm{ode}\left[1\right]\,\mathrm{way}=2\right)$

$\left[\mathrm{\_\ξ}=0\,\mathrm{\_\η}=\frac{y^3{x}^2}{2x^2{y}^3-1}\right],\left[\mathrm{\_\ξ}=0\,\mathrm{\_\η}=\frac{y\left(4x^3{y}^2+4x^2{y}^3+1\right)}{4\left(2x^2{y}^3-1\right)}\right]$

$\mathrm{dsolve}\left(\mathrm{ode}\left[1\right]\,\mathrm{way}=2\,\mathrm{fat}\,\mathrm{implicit}\right)$

$-2x-\frac{1}{2{y\left(x\right)}^2{x}^2}-2y\left(x\right)+\mathrm{c\_\_1}=0$

$\mathrm{ode}\left[2\right]≔\left(x{y\left(x\right)}^4+2x^2{y\left(x\right)}^3+2y\left(x\right)+x\right)\mathrm{diff}\left(y\left(x\right)\,x\right)+{y\left(x\right)}^5+y\left(x\right)$

${\mathrm{ode}}_2≔\left(x{y\left(x\right)}^4+2x^2{y\left(x\right)}^3+2y\left(x\right)+x\right)\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)+{y\left(x\right)}^5+y\left(x\right)$

$\mathrm{odeadvisor}\left(\mathrm{ode}\left[2\right]\right)$

$\left[\mathrm{\_rational}\right]$

$\mathrm{symgen}\left(\mathrm{ode}\left[2\right]\,\mathrm{way}=2\right)$

$\left[\mathrm{\_\ξ}=0\,\mathrm{\_\η}=\frac{y^2{\left(x+y\right)}^2}{2x^2{y}^3+x{y}^4+x+2y}\right],\left[\mathrm{\_\ξ}=0\,\mathrm{\_\η}=\frac{y\left(x+y\right)\left(x{y}^3-1\right)}{2x^2{y}^3+x{y}^4+x+2y}\right]$

$\mathrm{dsolve}\left(\mathrm{ode}\left[2\right]\,\mathrm{way}=2\,\mathrm{fat}\,\mathrm{implicit}\right)$

$\frac{{y\left(x\right)}^4+1}{y\left(x\right)\left(x+y\left(x\right)\right)}-{y\left(x\right)}^2+\mathrm{c\_\_1}=0$