Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$(y^{2} − 1) + 2(x − y(1 + y)^{2})y' = 0$ 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 ??

share|cite|improve this question
What about making this ODE exact by multiplying it by a proper integrating factor? – Applied mathematician Nov 19 '12 at 11:21 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.

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





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





share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.