 separablesol - Maple Help

DEtools

 separablesol
 find solutions of a separable first order ODE Calling Sequence separablesol(lode, v) Parameters

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

 • The separablesol routine determines whether the first argument is a separable 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 variable in the differential equation.
 • This function is part of the DEtools package, and so it can be used in the form separablesol(..) only after executing the command with(DEtools). However, it can always be accessed through the long form of the command by using DEtools[separablesol](..). Examples

 > $\mathrm{with}\left(\mathrm{DEtools}\right):$
 > $\mathrm{ode}≔{t}^{2}\left(z\left(t\right)+1\right)+{z\left(t\right)}^{2}\left(t-1\right)\left(\frac{ⅆ}{ⅆt}z\left(t\right)\right)=0$
 ${\mathrm{ode}}{≔}{{t}}^{{2}}{}\left({z}{}\left({t}\right){+}{1}\right){+}{{z}{}\left({t}\right)}^{{2}}{}\left({t}{-}{1}\right){}\left(\frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{z}{}\left({t}\right)\right){=}{0}$ (1)
 > $\mathrm{separablesol}\left(\mathrm{ode},z\left(t\right)\right)$
 $\left\{{z}{}\left({t}\right){=}{{ⅇ}}^{{\mathrm{RootOf}}{}\left({{ⅇ}}^{{2}{}{\mathrm{_Z}}}{+}{{t}}^{{2}}{-}{4}{}{{ⅇ}}^{{\mathrm{_Z}}}{+}{2}{}{\mathrm{ln}}{}\left({t}{-}{1}\right){+}{2}{}\mathrm{c__1}{+}{2}{}{\mathrm{_Z}}{+}{2}{}{t}{+}{3}\right)}{-}{1}\right\}$ (2)
 > $\mathrm{ode}≔t\left(z\left(t\right)-3\right)\mathrm{D}\left(z\right)\left(t\right)+4z\left(t\right)=0$
 ${\mathrm{ode}}{≔}{t}{}\left({z}{}\left({t}\right){-}{3}\right){}{\mathrm{D}}{}\left({z}\right){}\left({t}\right){+}{4}{}{z}{}\left({t}\right){=}{0}$ (3)
 > $\mathrm{separablesol}\left(\mathrm{ode},z\left(t\right)\right)$
 $\left\{{z}{}\left({t}\right){=}{{ⅇ}}^{{-}{\mathrm{LambertW}}{}\left({-}\frac{{{t}}^{{4}}{{3}}}{}{{ⅇ}}^{\frac{{4}{}\mathrm{c__1}}{{3}}}}{{3}}\right){+}\frac{{4}{}{\mathrm{ln}}{}\left({t}\right)}{{3}}{+}\frac{{4}{}\mathrm{c__1}}{{3}}}\right\}$ (4)