0
$\begingroup$

$$ \omega_\xi - \frac{\omega}{{\xi}} = \xi e^{\xi} $$

I dont understand how to get the integrating factor for this equation,

the answer is $ \frac{1}{\xi} $

how to obtain this? someone please show each step.

thanks

$\endgroup$
  • $\begingroup$ What should \omegasub be? do you want $\omega_\xi$ (\omega_\xi)? $\endgroup$ – AlexR Jun 13 '15 at 16:44
  • $\begingroup$ yes thats what i want, sorry $\endgroup$ – italy Jun 13 '15 at 16:45
  • $\begingroup$ No problem. I fixed it for you. $\endgroup$ – AlexR Jun 13 '15 at 16:46
1
$\begingroup$

We have

$$\omega' +\color{blue}{(-1/\xi)}\omega = \xi e^\xi \quad\text{ where } \ ' = {d\ \over d\xi}$$

Then the integrating factor is $$\exp\left(\int \color{blue}{(-1/\xi)} \ d\xi\right) = \exp(-\ln(\xi)) = \boxed{1 \over \xi}$$

Thus

$$\frac{\omega'}{\xi} - \frac{\omega}{\xi^2} = e^\xi$$

or

$$\left(\frac{\omega}{\xi}\right)' = e^\xi$$

$\endgroup$
  • $\begingroup$ OKay, i understand now, thank you. $\endgroup$ – italy Jun 13 '15 at 17:13

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.