Linearization of Ricatti equations

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In a Riccati equation y'=qo+q1y+q2y^2, if q2 is nonzero then you can make a substitution
v=yq2, S=q2q0, R=q1+(q2'/q2) which satisfies a Riccati, v'=v^2+R(x)v+S(x)=(yq2)'=q0q2+(q1+q2'/q2)v+v^2
with double substitution v=-u'/u, u now satisfies a linear 2nd ODE:
u''-R(x)u'+S(x)u=0, v'=-(u'/u)'=-(u''/u)+v^2, u''/u=v^2-v'=-S+Ru'/u, therefore we have reduced the Ricatti to an equation
u''+Su-Ru'=0, solving this equation will lead to a solution y=-u'/(q2u)

were these substitutions simply invented or is there a reasoning process behind the proof?
 

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Apparently you can use Lie theory to understand when a Riccati equation is integrable. That link might help, but that's about all I know at present unfortunately.
 
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pasmith
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In a Riccati equation y'=qo+q1y+q2y^2, if q2 is nonzero then you can make a substitution
v=yq2, S=q2q0, R=q1+(q2'/q2) which satisfies a Riccati, v'=v^2+R(x)v+S(x)=(yq2)'=q0q2+(q1+q2'/q2)v+v^2
with double substitution v=-u'/u, u now satisfies a linear 2nd ODE:
u''-R(x)u'+S(x)u=0, v'=-(u'/u)'=-(u''/u)+v^2, u''/u=v^2-v'=-S+Ru'/u, therefore we have reduced the Ricatti to an equation
u''+Su-Ru'=0, solving this equation will lead to a solution y=-u'/(q2u)

were these substitutions simply invented or is there a reasoning process behind the proof?
Since [itex]q_2 \neq 0[/itex], we can simplify things by changing variables so that [itex]q_2 = 1[/itex].

If we start with [itex]y' = q_0 + q_1 y + q_2 y^2[/itex] and substitute [itex]v = fy[/itex], then we end up with
[tex]
v' = \frac{q_2}{f}v^2 + \left(q_1 + \frac{f'}{f}\right)v + fq_0
[/tex]
and we can make the coefficient of [itex]v^2[/itex] equal to 1 by taking [itex]f = q_2[/itex], so that we have
[tex]
v' = v^2 + Rv + S
[/tex]
where [itex]R = q_1 + q_2'/q_2[/itex] and [itex]S = q_2q_0[/itex]. Now we want to get rid of the [itex]v^2[/itex] entirely, and we might consider the substitution [itex]v = gh[/itex]. That yields
[tex]
g'h + h'g = g^2h^2 + Rgh + S.
[/tex]
Now we must eliminate the [itex]g^2h^2[/itex] term by setting it equal to either [itex]g'h[/itex] or [itex]h'g[/itex]. It is clear from the symmetry that doing one gives the same as doing the other with g and h swapped. So taking the second we have
[tex]
h'g = g^2h^2 \Rightarrow h' = gh^2 \Rightarrow g = \frac{h'}{h^2} = -\left(\frac1h\right)'.
[/tex]
The natural thing to do now is to set [itex]h = -1/u[/itex], so that [itex]g = u'[/itex] and [itex]v = -u'/u[/itex]. Substituting those back we find
[tex]
g'h = Rgh + S \Rightarrow -u''/u = -Ru'/u + S
[/tex]
and thus
[tex]
u'' - Ru' + Su = 0
[/tex]
which is linear.
 
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