 Bernoulli - Maple Help

Solving Bernoulli's ODEs Description

 • The general form of Bernoulli's equation is given by:
 > Bernoulli_ode := diff(y(x),x)+f(x)*y(x)+g(x)*y(x)^a;
 ${\mathrm{Bernoulli_ode}}{≔}\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){+}{f}{}\left({x}\right){}{y}{}\left({x}\right){+}{g}{}\left({x}\right){}{{y}{}\left({x}\right)}^{{a}}$ (1)
 where f(x) and g(x) are arbitrary functions, and a is a symbolic power. See Differentialgleichungen, by E. Kamke, p. 19. Basically, the method consists of making a change of variables, leading to a linear equation which can be solved in general manner. The transformation is given by the following: Examples

 > $\mathrm{with}\left(\mathrm{DEtools},\mathrm{odeadvisor}\right)$
 $\left[{\mathrm{odeadvisor}}\right]$ (2)
 > $\mathrm{odeadvisor}\left(\mathrm{Bernoulli_ode}\right)$
 $\left[{\mathrm{_Bernoulli}}\right]$ (3)
 > $\mathrm{with}\left(\mathrm{PDEtools},\mathrm{dchange}\right)$
 $\left[{\mathrm{dchange}}\right]$ (4)
 > $\mathrm{ITR}≔\left\{x=t,y\left(x\right)={u\left(t\right)}^{\frac{1}{1-a}}\right\}$
 ${\mathrm{ITR}}{≔}\left\{{x}{=}{t}{,}{y}{}\left({x}\right){=}{{u}{}\left({t}\right)}^{\frac{{1}}{{1}{-}{a}}}\right\}$ (5)

and the ODE becomes

 > $\mathrm{new_ode}≔\mathrm{dchange}\left(\mathrm{ITR},\mathrm{Bernoulli_ode},\left[u\left(t\right),t\right]\right):$
 > $\mathrm{new_ode2}≔\mathrm{solve}\left(\mathrm{new_ode},\left\{\mathrm{diff}\left(u\left(t\right),t\right)\right\}\right):$
 > $\mathrm{op}\left(\mathrm{factor}\left(\mathrm{combine}\left(\mathrm{expand}\left(\mathrm{new_ode2}\right),\mathrm{power}\right)\right)\right)$
 $\frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{u}{}\left({t}\right){=}\left({a}{-}{1}\right){}\left({{u}{}\left({t}\right)}^{\frac{{a}}{{a}{-}{1}}}{}{g}{}\left({t}\right){}{\left({{u}{}\left({t}\right)}^{{-}\frac{{1}}{{a}{-}{1}}}\right)}^{{a}}{+}{u}{}\left({t}\right){}{f}{}\left({t}\right)\right)$ (6)

This ODE can then be solved by dsolve. Afterwards, another change of variables will reintroduce the original variables x and y(x).

The present implementation of dsolve can arrive directly at a general solution for Bernoulli's equation:

 > $\mathrm{ans}≔\mathrm{dsolve}\left(\mathrm{Bernoulli_ode}\right)$
 ${\mathrm{ans}}{≔}{y}{}\left({x}\right){=}\frac{{{ⅇ}}^{\frac{{\int }{f}{}\left({x}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}}{{a}{-}{1}}}}{{\left({a}{}\left({\int }\frac{{{ⅇ}}^{{\int }{f}{}\left({x}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}}{}{g}{}\left({x}\right)}{{{ⅇ}}^{\left({\int }{f}{}\left({x}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}\right){}{a}}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}\right){+}{\mathrm{_C1}}{-}\left({\int }\frac{{{ⅇ}}^{{\int }{f}{}\left({x}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}}{}{g}{}\left({x}\right)}{{{ⅇ}}^{\left({\int }{f}{}\left({x}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}\right){}{a}}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}\right)\right)}^{\frac{{1}}{{a}{-}{1}}}{}{{ⅇ}}^{\frac{\left({\int }{f}{}\left({x}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}\right){}{a}}{{a}{-}{1}}}}$ (7)