Solving 1st Order ODE: f(s) = -F(s)ln(s^2 + 1)/2 + C

[link2001]

I need to solve the following for f(s):

(s^2 + 1)f '(s) + s f(s) = 0

First I isolated for f '(s), which gave me:

f '(s) = -s f(s)/(s^2 + 1)

Then,

d f(s)/ds = -s f(s)/(s^2 + 1)

so, d f(s) = (-s f(s)/(s^2 + 1))ds

Integrating I get:

f(s) = -F(s)ln(s^2 + 1)/2 + C, where F(s) is the antiderivative of f(s) and C is a constant of integration.

Did I do any of that correctly??!??

 

[saltydog]

Not the last part. Just write it this way:

f^{'}+\frac{s}{s^2+1}f=0

That's in standard form right? Now calculate the integrating factor \sigma, multiply both sides by it, end up with an exact differential on the LHS, integrate, don't forget the constant of integration, that should do it. This is the integrating factor:

\sigma=\text{Exp}\left[\int \frac{s}{(s^2+1)}\right]

Can you do the rest?