$(y^{2} − 1) + 2(x − y(1 + y)^{2})y' = 0$ What method should I use to solve this ODE?

$(y^{2} − 1) + 2(x − y(1 + y)^{2})y' = 0$ What method should I use to solve this ODE ? Clearly This is not linear, and I don't know how to convert this to Bernoulli,finding integration factor seems to be more complicated then the actual question ? Am I missing something ? I think is somewhere along the line $y' = F(y/x)$...but what is $F$

I think I have to do some kind of trig sub...any one can tell me how to transform a 1st Order Ode to polar coordinate ??

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What about making this ODE exact by multiplying it by a proper integrating factor? – Applied mathematician Nov 19 '12 at 11:21
Ok..how to figure out this integrating factor then ? making $y^2-1 = M$ the coefficient of $y'$ is $N$ $M_y = 2y,N_x = 2$, Then ? – Archer Nov 19 '12 at 11:44

As you did if we put $M(x,y)=y^2-1$ and $N(x,y)=2(x-y\big(1+y\big)^2)$ then $$M_y=2y,N_x=2$$ so according to well-know formula, if the integrating factor be respect to $y$, then $$\mu(y)=\exp\left(\int\frac{M_y-N_x}{-M}dy\right)=\exp\left(\int\frac{2y-2}{1-y^2}dy\right)=\exp\left(\int\frac{-2}{1+y}dy\right)=\exp\left(\ln\left(\frac{1}{(1+y)^2}\right)\right)=\frac{1}{(1+y)^2}$$ Now multiply that to both sides of the equation. It makes your equation exact.

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Wow, thx. I never know this. Where can I find out the derivation of this formula ? – Archer Nov 19 '12 at 12:24

$(y^2-1)+2(x-y(1+y)^2)y'=0$

$(y+1)(y-1)+(2x-2y(y+1)^2)\dfrac{dy}{dx}=0$

$\dfrac{dx}{dy}+\dfrac{2x}{(y+1)(y-1)}-\dfrac{2y(y+1)}{y-1}=0$

$\dfrac{dx}{dy}+\dfrac{2x}{(y+1)(y-1)}=\dfrac{2y(y+1)}{y-1}$

I.F. $=e^{\int\frac{2}{(y+1)(y-1)}dy}=e^{\int\left(\frac{1}{y-1}-\frac{1}{y+1}\right)dy}=e^{\ln(y-1)-\ln(y+1)}=e^{\ln\frac{y-1}{y+1}}=\dfrac{y-1}{y+1}$

$\therefore\dfrac{d}{dy}\left(\dfrac{(y-1)x}{y+1}\right)=2y$

$\dfrac{(y-1)x}{y+1}=\int2y~dy$

$\dfrac{(y-1)x}{y+1}=y^2+C$

$x=\dfrac{(y+1)(y^2+C)}{y-1}$

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