ODEs Having Linear Symmetries - Maple Help

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ODEs Having Linear Symmetries

Description

 • The general forms of ODEs having one of the following linear symmetries
 > [xi=a+b*x, eta=0], [xi=a+b*y, eta=0], [xi=0, eta=c+d*x], [xi=0, eta=c+d*y]:
 • where the infinitesimal symmetry generator is given by:
 > G := f -> xi*diff(f,x) + eta*diff(f,y);
 ${G}{:=}{f}{→}{\mathrm{ξ}}{}\left(\frac{{\partial }}{{\partial }{x}}{}{f}\right){+}{\mathrm{η}}{}\left(\frac{{\partial }}{{\partial }{y}}{}{f}\right)$ (1)
 •
 • are given by:
 > ode[1] := DEtools[equinv]([xi=a+b*x, eta=0], y(x), 2);
 ${{\mathrm{ode}}}_{{1}}{:=}\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}{}{y}{}\left({x}\right){=}\frac{{\mathrm{_F1}}{}\left({y}{}\left({x}\right){,}\left(\frac{{ⅆ}}{{ⅆ}{x}}{}{y}{}\left({x}\right)\right){}\left({b}{}{x}{+}{a}\right)\right)}{{\left({b}{}{x}{+}{a}\right)}^{{2}}}$ (2)
 > ode[2] := DEtools[equinv]([xi=a+b*y, eta=0], y(x), 2);
 ${{\mathrm{ode}}}_{{2}}{:=}\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}{}{y}{}\left({x}\right){=}\frac{{\mathrm{_F1}}{}\left({y}{}\left({x}\right){,}{-}\frac{\left(\frac{{ⅆ}}{{ⅆ}{x}}{}{y}{}\left({x}\right)\right){}{b}{}{x}{-}{b}{}{y}{}\left({x}\right){-}{a}}{\left(\frac{{ⅆ}}{{ⅆ}{x}}{}{y}{}\left({x}\right)\right){}\left({b}{}{y}{}\left({x}\right){+}{a}\right)}\right){}{\left(\frac{{ⅆ}}{{ⅆ}{x}}{}{y}{}\left({x}\right)\right)}^{{3}}}{{\left({b}{}{y}{}\left({x}\right){+}{a}\right)}^{{3}}}$ (3)
 > ode[3] := DEtools[equinv]([xi=0, eta=c+d*x], y(x), 2);
 ${{\mathrm{ode}}}_{{3}}{:=}\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}{}{y}{}\left({x}\right){=}{\mathrm{_F1}}{}\left({x}{,}\frac{\left(\frac{{ⅆ}}{{ⅆ}{x}}{}{y}{}\left({x}\right)\right){}{d}{}{x}{+}\left(\frac{{ⅆ}}{{ⅆ}{x}}{}{y}{}\left({x}\right)\right){}{c}{-}{y}{}\left({x}\right){}{d}}{{d}{}{x}{+}{c}}\right)$ (4)
 > ode[4] := DEtools[equinv]([xi=0, eta=c+d*y], y(x), 2);
 ${{\mathrm{ode}}}_{{4}}{:=}\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}{}{y}{}\left({x}\right){=}{\mathrm{_F1}}{}\left({x}{,}\frac{\frac{{ⅆ}}{{ⅆ}{x}}{}{y}{}\left({x}\right)}{{y}{}\left({x}\right){}{d}{+}{c}}\right){}{d}{}{y}{}\left({x}\right){+}{\mathrm{_F1}}{}\left({x}{,}\frac{\frac{{ⅆ}}{{ⅆ}{x}}{}{y}{}\left({x}\right)}{{y}{}\left({x}\right){}{d}{+}{c}}\right){}{c}$ (5)
 Although the symmetries of these families of ODEs can be determined in a direct manner (using symgen), the simplicity of their pattern motivated us to have separate routines for recognizing them.

Examples

 > $\mathrm{with}\left(\mathrm{DEtools},\mathrm{equinv},\mathrm{odeadvisor},\mathrm{symgen}\right):$
 > $\mathrm{odeadvisor}\left({\mathrm{ode}}_{1}\right)$
 $\left[\left[{\mathrm{_2nd_order}}{,}{\mathrm{_with_linear_symmetries}}\right]\right]$ (6)
 > $\mathrm{odeadvisor}\left({\mathrm{ode}}_{2}\right)$
 $\left[\left[{\mathrm{_2nd_order}}{,}{\mathrm{_with_linear_symmetries}}\right]\right]$ (7)
 > $\mathrm{odeadvisor}\left({\mathrm{ode}}_{3}\right)$
 $\left[\left[{\mathrm{_2nd_order}}{,}{\mathrm{_with_linear_symmetries}}\right]\right]$ (8)
 > $\mathrm{odeadvisor}\left({\mathrm{ode}}_{4}\right)$
 $\left[\left[{\mathrm{_2nd_order}}{,}{\mathrm{_with_linear_symmetries}}\right]\right]$ (9)

As an example that can be solved by the related routine, consider

 > ${\mathrm{ode}}_{5}≔\mathrm{equinv}\left(\left[\left[0,y\right],\left[x,0\right]\right],y\left(x\right),2\right)$
 ${{\mathrm{ode}}}_{{5}}{:=}\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}{}{y}{}\left({x}\right){=}\frac{{\mathrm{_F1}}{}\left(\frac{\left(\frac{{ⅆ}}{{ⅆ}{x}}{}{y}{}\left({x}\right)\right){}{x}}{{y}{}\left({x}\right)}\right){}{y}{}\left({x}\right)}{{{x}}^{{2}}}$ (10)
 > $\mathrm{dsolve}\left({\mathrm{ode}}_{5}\right)$
 ${y}{}\left({x}\right){=}{{ⅇ}}^{{{∫}}_{{}}^{{\mathrm{ln}}{}\left({x}\right)}{\mathrm{RootOf}}{}\left({{∫}}_{{}}^{{\mathrm{_Z}}}\frac{{1}}{{\mathrm{_a}}{-}{{\mathrm{_a}}}^{{2}}{+}{\mathrm{_F1}}{}\left({\mathrm{_a}}\right)}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{\mathrm{_a}}{-}{\mathrm{_b}}{+}{\mathrm{_C1}}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{\mathrm{_b}}{+}{\mathrm{_C2}}}$ (11)