Question

What is the difference between bernoulii equation and riccati equation

Answer 1

The Bernoulli equation is almost identical to the standard linear ODE.

$y'=P(x)y+Q(x)y^n$

Compare to the basic linear ODE,

$y'=P(x)y+Q(x)$

Meanwhile, the Riccati equation takes the form

$y'=P(x)+Q(x)y+R(x)y^2$

which in special cases is of Bernoulli type if $P(x)=0$, and linear if $R(x)=0$. But in general each type takes a different method to solve. From now on, I'll abbreviate the coefficient functions as $P,Q,R$ for brevity.

For Bernoulli equations, the standard approach is to write

$y'=Py+Qy^n$
$y^{-n}y'=Py^{1-n}+Q$

and substitute $v=y^{1-n}$. This makes $v'=(1-n)y^{-n}y'$, so the ODE is rewritten as

$\dfrac1{1-n}v'=Pv+Q$

and the equation is now linear in $v$.

The Riccati equation, on the other hand, requires a different substitution. Set $v=Ry$, so that $v'=R'y+Ry'=R'\dfrac vR+Ry'$. Then you have

$y'=P+Qy+Ry^2$
$\dfrac{v'-R'\dfrac vR}R=P+Q\dfrac vR+R\dfrac{v^2}{R^2}$
$v'=PR+\left(Q+\dfrac{R'}R\right)v+v^2$

Next, setting $v=\dfrac{u'}u$, so that $v'=\dfrac{uu''-(u')^2}{u^2}$, allows you to write this as a linear second-order equation. You have

$\dfrac{uu''-(u')^2}{u^2}=PR+\left(Q+\dfrac{R'}R\right)\dfrac{u'}u+\dfrac{(u')^2}{u^2}$
$u''-\left(Q+\dfrac{R'}R\right)u'+PRu=0$
$u''+Su'+Tu=0$

where $S=-\left(Q+\dfrac{R'}R\right)$ and $T=PR$.