Show that the Riccati equation boils down to linear form in v

**Arturo_026** · Mar 14th 2011, 05:23 PM

I've been trying this problem but in both ways that i try it i get stuck. i would like to get any hints onto how to start it.

An equation: dy/dx = P(x)y^2 + Q(x)y + R(x) is called the riccati form...

If one solution, say u(x), of this is known, show that the substitution y=u+(1/v) reduces the riccati equation into a linear equation in v.

Thnak u very much!

**Ackbeet** · Mar 14th 2011, 05:41 PM

Well, what have you tried so far?

**Arturo_026** · Mar 15th 2011, 04:24 PM

First I start with y=u+(1/v) and from this I get: dy/dx = u' - (1/v^2)(dv/dx)

Then I put these last expression into the given riccati form.

From here I tried 2 things:
1) In the riccati form I substituded y by u(x)... this led me to call P(x)u^2 + Q(x)u + R(x) = -F(x) ... and u' = G(x)
From here I have dv/dx + v^2G(x) = v^2F(x) ... The problem is that this is not linear, and not even bernuolli seems to be available.

2) I substitute y=u+(1/v) in the riccati form, but it gets super ugly and it doesnt seem to go anywhere.

Thanks!

**Ackbeet** · Mar 15th 2011, 07:16 PM

Hmm. You have $y'=P(x)\,y^{2}+Q(x)\,y+R(x),$ with $u$ a known particular solution. Your goal is to show that the substitution $y=u+1/v$ renders the equation linear in $v.$ Your derivative is correct. Now plug everything into the DE:

$u'-\dfrac{1}{v^{2}}\,\dfrac{dv}{dx}=P(x)\left(u^{2}+\ dfrac{2u}{v}+\dfrac{1}{v^{2}}\right)+Q(x)\left(u+\ dfrac{1}{v}\right)+R(x).$

Rearranging a bit yields

$\underbrace{u'-P(x)u^{2}-Q(x)u-R(x)}_{\text{=0, since u is a particular solution.}}=\dfrac{1}{v^{2}}\,\dfrac{dv}{dx}+P(x)\ left(\dfrac{2u}{v}+\dfrac{1}{v^{2}}\right)+Q(x)\df rac{1}{v}.$

Can you finish from here?

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