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?