1. ## Linear Differential Equation

dy/dx + 2y/x = (cosx)/x

So far I have:

P(x) = 2/x
Q(x) = (cosx)/x
Integrating Factor: e^2lnx

So I times both sides by the integrating factor and get:

e^2lnx * y' + e^2lnx * 2y/x = e^2lnx * (cosx)/x

Now when I integrate both sides, how am I supposed to integrate (cosx)/x without using Taylor series?

2. ## Re: Linear Differential Equation

Can't you simplify $\displaystyle e^{2\ln x}$ using logarithmic identities?

3. ## Re: Linear Differential Equation

Oh of course! Thank you so much!