# integrating factor for differential equation

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

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

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

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

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

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