 rational - Maple Help

Solving Rational ODEs Description

 • 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));
 ${\mathrm{rational_ode}}{≔}\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){=}\frac{{\mathrm{P1}}{}\left({x}{,}{y}{}\left({x}\right)\right)}{{\mathrm{P2}}{}\left({x}{,}{y}{}\left({x}\right)\right)}$ (1)
 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). Examples

 > $\mathrm{with}\left(\mathrm{DEtools},\mathrm{odeadvisor},\mathrm{symgen}\right)$
 $\left[{\mathrm{odeadvisor}}{,}{\mathrm{symgen}}\right]$ (2)
 > $\mathrm{ode}\left[1\right]≔\left(2{x}^{3}{y\left(x\right)}^{3}-x\right)\mathrm{diff}\left(y\left(x\right),x\right)+2{x}^{3}{y\left(x\right)}^{3}-y\left(x\right)$
 ${{\mathrm{ode}}}_{{1}}{≔}\left({2}{}{{x}}^{{3}}{}{{y}{}\left({x}\right)}^{{3}}{-}{x}\right){}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){+}{2}{}{{x}}^{{3}}{}{{y}{}\left({x}\right)}^{{3}}{-}{y}{}\left({x}\right)$ (3)
 > $\mathrm{odeadvisor}\left(\mathrm{ode}\left[1\right]\right)$
 $\left[{\mathrm{_rational}}\right]$ (4)

Rational symmetries for ode

 > $\mathrm{symgen}\left(\mathrm{ode}\left[1\right],\mathrm{way}=2\right)$
 $\left[{\mathrm{_ξ}}{=}{0}{,}{\mathrm{_η}}{=}\frac{{{y}}^{{3}}{}{{x}}^{{2}}}{{2}{}{{y}}^{{3}}{}{{x}}^{{2}}{-}{1}}\right]{,}\left[{\mathrm{_ξ}}{=}{0}{,}{\mathrm{_η}}{=}\frac{{y}{}\left({4}{}{{x}}^{{3}}{}{{y}}^{{2}}{+}{4}{}{{y}}^{{3}}{}{{x}}^{{2}}{+}{1}\right)}{{2}{}{{y}}^{{3}}{}{{x}}^{{2}}{-}{1}}\right]$ (5)

Once you know 'the way' to find the symmetries, you can directly specify that dsolve should use this 'way'. Also, ODEs with rational symmetries are usually better integrated by using their symmetries to build an integrating factor (see intfactor) instead of using canonical coordinates. To avoid multiple solutions, you can request an implicit answer, as follows:

 > $\mathrm{dsolve}\left(\mathrm{ode}\left[1\right],\mathrm{way}=2,\mathrm{fat},\mathrm{implicit}\right)$
 ${-}{2}{}{x}{-}\frac{{1}}{{2}{}{{y}{}\left({x}\right)}^{{2}}{}{{x}}^{{2}}}{-}{2}{}{y}{}\left({x}\right){+}{\mathrm{_C1}}{=}{0}$ (6)
 > $\mathrm{ode}\left[2\right]≔\left(x{y\left(x\right)}^{4}+2{x}^{2}{y\left(x\right)}^{3}+2y\left(x\right)+x\right)\mathrm{diff}\left(y\left(x\right),x\right)+{y\left(x\right)}^{5}+y\left(x\right)$
 ${{\mathrm{ode}}}_{{2}}{≔}\left({x}{}{{y}{}\left({x}\right)}^{{4}}{+}{2}{}{{x}}^{{2}}{}{{y}{}\left({x}\right)}^{{3}}{+}{2}{}{y}{}\left({x}\right){+}{x}\right){}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){+}{{y}{}\left({x}\right)}^{{5}}{+}{y}{}\left({x}\right)$ (7)
 > $\mathrm{odeadvisor}\left(\mathrm{ode}\left[2\right]\right)$
 $\left[{\mathrm{_rational}}\right]$ (8)
 > $\mathrm{symgen}\left(\mathrm{ode}\left[2\right],\mathrm{way}=2\right)$
 $\left[{\mathrm{_ξ}}{=}{0}{,}{\mathrm{_η}}{=}\frac{{{y}}^{{2}}{}{\left({x}{+}{y}\right)}^{{2}}}{{2}{}\left({2}{}{{y}}^{{3}}{}{{x}}^{{2}}{+}{x}{}{{y}}^{{4}}{+}{x}{+}{2}{}{y}\right)}\right]{,}\left[{\mathrm{_ξ}}{=}{0}{,}{\mathrm{_η}}{=}\frac{{y}{}\left({x}{+}{y}\right){}\left({{y}}^{{3}}{}{x}{-}{1}\right)}{{2}{}{{y}}^{{3}}{}{{x}}^{{2}}{+}{x}{}{{y}}^{{4}}{+}{x}{+}{2}{}{y}}\right]$ (9)
 > $\mathrm{dsolve}\left(\mathrm{ode}\left[2\right],\mathrm{way}=2,\mathrm{fat},\mathrm{implicit}\right)$
 ${\mathrm{_C1}}{+}\frac{{{y}{}\left({x}\right)}^{{4}}{+}{1}}{{y}{}\left({x}\right){}\left({x}{+}{y}{}\left({x}\right)\right)}{-}{{y}{}\left({x}\right)}^{{2}}{=}{0}$ (10)