genhomosol - Maple Help

DEtools

 genhomosol
 find solutions of a homogeneous first order ODE

 Calling Sequence genhomosol(lode, v)

Parameters

 ode - first order differential equation v - dependent variable of the lode

Description

 • The genhomosol routine determines whether the first argument is a general homogeneous first order ODE and, if so, returns a solution to the equation.
 • The first argument is a differential equation in diff or D form and the second argument is the function in the differential equation.
 • This function is part of the DEtools package, and so it can be used in the form genhomosol(..) only after executing the command with(DEtools). However, it can always be accessed through the long form of the command by using DEtools[genhomosol](..).

Examples

 > $\mathrm{with}\left(\mathrm{DEtools}\right):$
 > $\mathrm{ode}≔{t}^{3}+{z\left(t\right)}^{3}-3t{z\left(t\right)}^{2}\left(\frac{ⅆ}{ⅆt}z\left(t\right)\right)=0:$
 > $\mathrm{genhomosol}\left(\mathrm{ode},z\left(t\right)\right)$
 $\left\{{z}{}\left({t}\right){=}\frac{{{4}}^{{1}}{{3}}}{}{\left({t}{}\left({\mathrm{c__1}}^{{2}}{}{{t}}^{{2}}{+}{1}\right){}\mathrm{c__1}\right)}^{{1}}{{3}}}}{{2}{}\mathrm{c__1}}{,}{z}{}\left({t}\right){=}{-}\frac{{{4}}^{{1}}{{3}}}{}{\left({t}{}\left({\mathrm{c__1}}^{{2}}{}{{t}}^{{2}}{+}{1}\right){}\mathrm{c__1}\right)}^{{1}}{{3}}}}{{4}{}\mathrm{c__1}}{-}\frac{{I}{}\sqrt{{3}}{}{{4}}^{{1}}{{3}}}{}{\left({t}{}\left({\mathrm{c__1}}^{{2}}{}{{t}}^{{2}}{+}{1}\right){}\mathrm{c__1}\right)}^{{1}}{{3}}}}{{4}{}\mathrm{c__1}}{,}{z}{}\left({t}\right){=}{-}\frac{{{4}}^{{1}}{{3}}}{}{\left({t}{}\left({\mathrm{c__1}}^{{2}}{}{{t}}^{{2}}{+}{1}\right){}\mathrm{c__1}\right)}^{{1}}{{3}}}}{{4}{}\mathrm{c__1}}{+}\frac{{I}{}\sqrt{{3}}{}{{4}}^{{1}}{{3}}}{}{\left({t}{}\left({\mathrm{c__1}}^{{2}}{}{{t}}^{{2}}{+}{1}\right){}\mathrm{c__1}\right)}^{{1}}{{3}}}}{{4}{}\mathrm{c__1}}\right\}$ (1)
 > $\mathrm{ode}≔-z\left(t\right)-\sqrt{{t}^{2}-{z\left(t\right)}^{2}}+t\mathrm{D}\left(z\right)\left(t\right)=0:$
 > $\mathrm{genhomosol}\left(\mathrm{ode},z\left(t\right)\right)$
 $\left\{{-}{\mathrm{arctan}}{}\left(\frac{{z}{}\left({t}\right)}{\sqrt{{{t}}^{{2}}{-}{{z}{}\left({t}\right)}^{{2}}}}\right){+}{\mathrm{ln}}{}\left({t}\right){-}\mathrm{c__1}{=}{0}\right\}$ (2)