# Solving second order linear ordinary differential equation

I am having trouble solving this differential equation:

$$y'' + 2(x-x^{3})^{-1}y'=0\\$$

do I find the integrating factor:

$$u= e^{\int 2(x-x^3)^{-1}}\\$$

or is there another method?

This differential equation is part of a larger problem on the reduction of order method, so if this equation I presented here does not make sense or is insolvable by traditional ways, I will include the whole problem.

The included solution, it is $$y_1(x)=x\\$$

• I think integrating factor is the way to go. Well, $\frac{1}{x-x^3} = \frac{1}{x(1-x^2)}$, partial fraction decomposition maybe? – Chee Han Feb 13 '17 at 4:02
• @BryanChen: You could try letting $y' = v$. – Moo Feb 13 '17 at 4:02
• By the way you did mention reduction of order method, are you given a known solution to the differential equation? – Chee Han Feb 13 '17 at 4:02
• write $\dfrac{dv}{v}=-2(x-x^3)^{-1}dx$ after BryanChen Changing – Nosrati Feb 13 '17 at 4:03
• Sorry, I should've given the included solution, it is $$y_1(x)=x\\$$ @Chee Han – Bryan Chen Feb 13 '17 at 4:38