2
$\begingroup$

Does anybody suggest how to face the inhomogeneous Bernoulli differential equation $y'+P(x)y=Q(x)y^n+f(x)$ for the simple case $f=const.$ and for the generic case. I would like to know about techniques of approximation, bounds, asymptotic limit, numerical techniques etc. Thank you Roberto

$\endgroup$
1

2 Answers 2

2
$\begingroup$

I presume your $f(t)$ should be $f(x)$. This is Chini's equation. See e.g. http://www.maplesoft.com/support/help/Maple/view.aspx?path=odeadvisor/Chini

$\endgroup$
1
  • $\begingroup$ Thank you, of course $f(t)$ is $f(x)$. Thanks for the answer. Roberto $\endgroup$ Aug 6, 2012 at 18:06
-2
$\begingroup$

Solution: let u=y^(1-n) du/dx = (1-n) y^(-n) dy/dx = (1-n) y^(-n) [ -Py+Qy^n+f] =-(1-n) Pu +(1-n)Q +(1-n)f u^(n/n-1)

Separation of variables [1/(1-n)]Int[du/ { f u^(n/n-1)+Pu-Q}]= Int[dx]= x+ C Factor out the division and integrates the LHS. We get F(u,x)=0 then transform into F(y,x)=0.

For n=2, it is very easy to solve. You can try it.

$\endgroup$
5
  • 2
    $\begingroup$ You can't separate the variables since $P,~Q,~f$ depend on $x$. Have you tried it? In general these equations are not solvable by quadrature when $PQf\neq0$. $\endgroup$ Nov 9, 2015 at 18:42
  • $\begingroup$ Yes, it can be done for the cases that PQf are constants. If they are functions of x, cannot solved by separation. Thks a lot for your reminding. I walked too quickly in simplified form. $\endgroup$ Nov 9, 2015 at 19:13
  • $\begingroup$ When PQf are constants and n=2, LHS becomes (1/1-n) (1/f)(1/(u1-u2) ln |(u-u1)/(u-u2)|= x+C where u1 and u2 are two roots of u^2 +P/f u - Q/f =0. $\endgroup$ Nov 9, 2015 at 19:29
  • $\begingroup$ When n=2, PQf depend on x. This is Ricatti's Eq. It can also be transformed to 2nd-order linear ODE. Seehttps://en.m.wikipedia.org/wiki/Riccati_equation $\endgroup$ Nov 9, 2015 at 19:45
  • 2
    $\begingroup$ Well, yes, I know that (and a lot of other people here also do…) so that referring to Wikipedia is besides the point. Just check your maths before posting, please. And also, please use LaTeX typesetting… $\endgroup$ Nov 9, 2015 at 20:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.