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;

$\mathrm{Bernoulli\_ode}≔\frac{\ⅆ}{\ⅆx}y\left(x\right)+f\left(x\right)y\left(x\right)+g\left(x\right){y\left(x\right)}^a$

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:

$\mathrm{with}\left(\mathrm{DEtools}\,\mathrm{odeadvisor}\right)$

$\left[\mathrm{odeadvisor}\right]$

$\mathrm{odeadvisor}\left(\mathrm{Bernoulli\_ode}\right)$

$\left[\mathrm{\_Bernoulli}\right]$

$\mathrm{with}\left(\mathrm{PDEtools}\,\mathrm{dchange}\right)$

$\left[\mathrm{dchange}\right]$

$\mathrm{ITR}≔\left\{x=t\,y\left(x\right)={u\left(t\right)}^{\frac{1}{1-a}}\right\}$

$\mathrm{ITR}≔\left\{x=t\,y\left(x\right)={u\left(t\right)}^{\frac{1}{1-a}}\right\}$

$\mathrm{new\_ode}≔\mathrm{dchange}\left(\mathrm{ITR}\,\mathrm{Bernoulli\_ode}\,\left[u\left(t\right)\,t\right]\right)\:$

$\mathrm{new\_ode2}≔\mathrm{solve}\left(\mathrm{new\_ode}\,\left\{\mathrm{diff}\left(u\left(t\right)\,t\right)\right\}\right)\:$

$\mathrm{op}\left(\mathrm{factor}\left(\mathrm{combine}\left(\mathrm{expand}\left(\mathrm{new\_ode2}\right)\,\mathrm{power}\right)\right)\right)$

$\frac{\ⅆ}{\ⅆt}u\left(t\right)=\left(-1+a\right)\left(g\left(t\right){u\left(t\right)}^{\frac{a}{-1+a}}{\left({u\left(t\right)}^{-\frac{1}{-1+a}}\right)}^a+u\left(t\right)f\left(t\right)\right)$

$\mathrm{ans}≔\mathrm{dsolve}\left(\mathrm{Bernoulli\_ode}\right)$

$\mathrm{ans}≔y\left(x\right)=\frac{{\ⅇ}^{\frac{\int f\left(x\right)\ⅆx}{-1+a}}}{{\left(a\left(\int \frac{{\ⅇ}^{\int f\left(x\right)\ⅆx}g\left(x\right)}{{\ⅇ}^{\left(\int f\left(x\right)\ⅆx\right)a}}\ⅆx\right)+\mathrm{c\_\_1}-\left(\int \frac{{\ⅇ}^{\int f\left(x\right)\ⅆx}g\left(x\right)}{{\ⅇ}^{\left(\int f\left(x\right)\ⅆx\right)a}}\ⅆx\right)\right)}^{\frac{1}{-1+a}}{\ⅇ}^{\frac{\left(\int f\left(x\right)\ⅆx\right)a}{-1+a}}}$