How to do $\int e^\frac{-1}{2x^2}x^{-5}dx$?

The original problem solve $xy'+\frac{y}{x^2}=\frac{1}{x^4}$

I simplified and found the integrating factor $e^\frac{-1}{2x^2}$ now the RHS of the equation is $\int e^\frac{-1}{2x^2}x^{-5}dx$

I have tried integration by parts with $u=e^\frac{-1}{2x^2}$ and $v=x^{-5}$ but this only seems to make it worse.

  • $\begingroup$ Try doing a $u$ substitution with $u=1/x$. $\endgroup$ – Sean Lake Oct 10 '17 at 3:26
  • 3
    $\begingroup$ $1/(2x^2)$ is probably the better $u$-substitution. $\endgroup$ – eyeballfrog Oct 10 '17 at 3:26

Let t = $1/2x^{2}$

dt = $(-x^{-3}dx)$

So integration converts to $\int {e^{-t}(-2t)dt}$ Now apply by parts

and then substitute back t.

  • $\begingroup$ It took a some time to see where you are going with this but it helped thank you $\endgroup$ – Gobabis Oct 10 '17 at 3:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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