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

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To solve the first-order ordinary differential equation (ODE) (s^2 + 1)f '(s) + s f(s) = 0, the user correctly isolated f '(s) and expressed it as d f(s)/ds = -s f(s)/(s^2 + 1). They integrated this to arrive at f(s) = -F(s)ln(s^2 + 1)/2 + C, where F(s) is the antiderivative of f(s). Another participant confirmed that the equation is in standard form and suggested calculating the integrating factor, which is σ = Exp[∫(s/(s^2 + 1))]. The discussion emphasizes the importance of using the integrating factor to solve the ODE correctly.
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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??!??
 
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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?
 
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