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
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1$\begingroup$ I fear you should move your question to math.stackexchange.com. $\endgroup$– JonCommented Aug 5, 2012 at 14:27
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2 Answers
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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
answered Aug 5, 2012 at 19:00
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$\begingroup$ Thank you, of course $f(t)$ is $f(x)$. Thanks for the answer. Roberto $\endgroup$ Commented Aug 6, 2012 at 18:06
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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.
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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 $PQf\neq0$. $\endgroup$ Commented Nov 9, 2015 at 18:42
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$\begingroup$ 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. $\endgroup$ Commented Nov 9, 2015 at 19:13
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$\begingroup$ 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. $\endgroup$ Commented Nov 9, 2015 at 19:29
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$\begingroup$ 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 $\endgroup$ Commented Nov 9, 2015 at 19:45
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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… $\endgroup$ Commented Nov 9, 2015 at 20:09