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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 $$y_1(x)=x\\$$

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    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? – 2017-02-13
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    @BryanChen: You could try letting $y' = v$. – 2017-02-13
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    By the way you did mention reduction of order method, are you given a known solution to the differential equation? – 2017-02-13
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    write $\dfrac{dv}{v}=-2(x-x^3)^{-1}dx$ after BryanChen Changing – 2017-02-13
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    Sorry, I should've given the included solution, it is $$y_1(x)=x\\$$ @Chee Han – 2017-02-13

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I am attaching the solution as an image file. Please find the attachment.

enter image description here

The integration you can find Here.

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    Why the downvote. If you don't want just tell we will not give the answers. – 2017-02-13
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    was not me who downvoted – 2017-02-13
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    Okay, this comment was for that who did it. – 2017-02-13