Solving Abel's ODEs of the Second Kind, Class A - Maple Programming Help

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Solving Abel's ODEs of the Second Kind, Class A

Description

 • The general form of Abel's equation, second kind, class A is given by:
 > Abel_ode2A := (y(x)+g(x))*diff(y(x),x)=f2(x)*y(x)^2+f1(x)*y(x)+f0(x);
 ${\mathrm{Abel_ode2A}}{≔}\left({y}{}\left({x}\right){+}{g}{}\left({x}\right)\right){}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){=}{\mathrm{f2}}{}\left({x}\right){}{{y}{}\left({x}\right)}^{{2}}{+}{\mathrm{f1}}{}\left({x}\right){}{y}{}\left({x}\right){+}{\mathrm{f0}}{}\left({x}\right)$ (1)
 where f2(x), f1(x), f0(x), and g(x) are arbitrary functions. See Differentialgleichungen, by E. Kamke, p. 26. There is as yet no general solution for this ODE.
 • Note that all ODEs of type Abel, second kind, can be rewritten as ODEs of type Abel, first kind, as explained in ?odeadvisor,Abel2C

Examples

 > $\mathrm{with}\left(\mathrm{DEtools},\mathrm{symgen},\mathrm{odeadvisor}\right)$
 $\left[{\mathrm{symgen}}{,}{\mathrm{odeadvisor}}\right]$ (2)
 > $\mathrm{odeadvisor}\left(\mathrm{Abel_ode2A}\right)$
 $\left[\left[{\mathrm{_Abel}}{,}{\mathrm{2nd type}}{,}{\mathrm{class A}}\right]\right]$ (3)

1) f0(x) = f1(x)*g(x)-f2(x)*g(x)^2

 > $\mathrm{ode}≔\mathrm{eval}\left(\mathrm{subs}\left(\mathrm{f0}\left(x\right)=\mathrm{f1}\left(x\right)g\left(x\right)-\mathrm{f2}\left(x\right){g\left(x\right)}^{2},\mathrm{Abel_ode2A}\right)\right)$
 ${\mathrm{ode}}{≔}\left({y}{}\left({x}\right){+}{g}{}\left({x}\right)\right){}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){=}{\mathrm{f2}}{}\left({x}\right){}{{y}{}\left({x}\right)}^{{2}}{+}{\mathrm{f1}}{}\left({x}\right){}{y}{}\left({x}\right){+}{g}{}\left({x}\right){}{\mathrm{f1}}{}\left({x}\right){-}{\mathrm{f2}}{}\left({x}\right){}{{g}{}\left({x}\right)}^{{2}}$ (4)

This case can be solved as follows:

 > $\mathrm{dsolve}\left(\mathrm{ode},y\left(x\right)\right)$
 ${y}{}\left({x}\right){=}{-}{g}{}\left({x}\right){,}{y}{}\left({x}\right){=}\left({\int }\left({-}{\mathrm{f2}}{}\left({x}\right){}{g}{}\left({x}\right){+}{\mathrm{f1}}{}\left({x}\right)\right){}{{ⅇ}}^{{-}\left({\int }{\mathrm{f2}}{}\left({x}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}\right)}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}{+}{\mathrm{_C1}}\right){}{{ⅇ}}^{{\int }{\mathrm{f2}}{}\left({x}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}}$ (5)

2) Another case which can be solved:

f1(x) = 2*f2(x)*g(x)-diff(g(x),x)

 > $\mathrm{ode}≔\mathrm{eval}\left(\mathrm{subs}\left(\mathrm{f1}\left(x\right)=2\mathrm{f2}\left(x\right)g\left(x\right)-\left(\frac{ⅆ}{ⅆx}g\left(x\right)\right),\mathrm{Abel_ode2A}\right)\right)$
 ${\mathrm{ode}}{≔}\left({y}{}\left({x}\right){+}{g}{}\left({x}\right)\right){}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){=}{\mathrm{f2}}{}\left({x}\right){}{{y}{}\left({x}\right)}^{{2}}{+}\left({2}{}{\mathrm{f2}}{}\left({x}\right){}{g}{}\left({x}\right){-}\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{g}{}\left({x}\right)\right){}{y}{}\left({x}\right){+}{\mathrm{f0}}{}\left({x}\right)$ (6)

Although the answer for this case can be obtained using standard methods (an integrating factor is easily found), the use of symmetry methods can provide an explicit solution. The infinitesimals for this case are given by

 > $\mathrm{symgen}\left(\mathrm{ode},y\left(x\right)\right)$
 $\left[{\mathrm{_ξ}}{=}{0}{,}{\mathrm{_η}}{=}\frac{{{ⅇ}}^{{\int }{2}{}{\mathrm{f2}}{}\left({x}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}}}{{y}{+}{g}{}\left({x}\right)}\right]$ (7)

To indicate the use of symmetry methods "at first", we can explicitly indicate an integration method (see dsolve); for instance, to use the canonical coordinates of the invariance group:

 > $\mathrm{ans}≔\mathrm{dsolve}\left(\mathrm{ode},y\left(x\right),\mathrm{can}\right)$
 ${\mathrm{ans}}{≔}{y}{}\left({x}\right){=}\frac{{-}{{ⅇ}}^{{-}{2}{}\left({\int }{\mathrm{f2}}{}\left({x}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}\right)}{}{g}{}\left({x}\right){+}\sqrt{{\left({{ⅇ}}^{{-}{2}{}\left({\int }{\mathrm{f2}}{}\left({x}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}\right)}\right)}^{{2}}{}{{g}{}\left({x}\right)}^{{2}}{+}{2}{}{{ⅇ}}^{{-}{2}{}\left({\int }{\mathrm{f2}}{}\left({x}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}\right)}{}\left({\int }\frac{{\mathrm{f0}}{}\left({x}\right)}{{\left({{ⅇ}}^{{\int }{\mathrm{f2}}{}\left({x}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}}\right)}^{{2}}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}\right){+}{2}{}{{ⅇ}}^{{-}{2}{}\left({\int }{\mathrm{f2}}{}\left({x}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}\right)}{}{\mathrm{_C1}}}}{{{ⅇ}}^{{-}{2}{}\left({\int }{\mathrm{f2}}{}\left({x}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}\right)}}{,}{y}{}\left({x}\right){=}{-}\frac{{{ⅇ}}^{{-}{2}{}\left({\int }{\mathrm{f2}}{}\left({x}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}\right)}{}{g}{}\left({x}\right){+}\sqrt{{\left({{ⅇ}}^{{-}{2}{}\left({\int }{\mathrm{f2}}{}\left({x}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}\right)}\right)}^{{2}}{}{{g}{}\left({x}\right)}^{{2}}{+}{2}{}{{ⅇ}}^{{-}{2}{}\left({\int }{\mathrm{f2}}{}\left({x}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}\right)}{}\left({\int }\frac{{\mathrm{f0}}{}\left({x}\right)}{{\left({{ⅇ}}^{{\int }{\mathrm{f2}}{}\left({x}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}}\right)}^{{2}}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}\right){+}{2}{}{{ⅇ}}^{{-}{2}{}\left({\int }{\mathrm{f2}}{}\left({x}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}\right)}{}{\mathrm{_C1}}}}{{{ⅇ}}^{{-}{2}{}\left({\int }{\mathrm{f2}}{}\left({x}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}\right)}}$ (8)