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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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  • $\begingroup$ 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? $\endgroup$ – Chee Han Feb 13 '17 at 4:02
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    $\begingroup$ @BryanChen: You could try letting $y' = v$. $\endgroup$ – Moo Feb 13 '17 at 4:02
  • $\begingroup$ By the way you did mention reduction of order method, are you given a known solution to the differential equation? $\endgroup$ – Chee Han Feb 13 '17 at 4:02
  • $\begingroup$ write $\dfrac{dv}{v}=-2(x-x^3)^{-1}dx$ after BryanChen Changing $\endgroup$ – Nosrati Feb 13 '17 at 4:03
  • $\begingroup$ Sorry, I should've given the included solution, it is $$y_1(x)=x\\$$ @Chee Han $\endgroup$ – Bryan Chen Feb 13 '17 at 4:38
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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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  • $\begingroup$ Why the downvote. If you don't want just tell we will not give the answers. $\endgroup$ – Sachchidanand Prasad Feb 13 '17 at 4:57
  • $\begingroup$ was not me who downvoted $\endgroup$ – Bryan Chen Feb 13 '17 at 5:11
  • $\begingroup$ Okay, this comment was for that who did it. $\endgroup$ – Sachchidanand Prasad Feb 13 '17 at 13:57

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