• Support PF! Buy your school textbooks, materials and every day products Here!

A first order ODE

  • Thread starter link2001
  • Start date
  • #1
8
0
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??!!??
 

Answers and Replies

  • #2
saltydog
Science Advisor
Homework Helper
1,582
2
Not the last part. Just write it this way:

[tex]f^{'}+\frac{s}{s^2+1}f=0[/tex]

That's in standard form right? Now calculate the integrating factor [itex]\sigma [/itex], 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:

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

Can you do the rest?
 

Related Threads for: A first order ODE

  • Last Post
Replies
10
Views
1K
  • Last Post
Replies
8
Views
3K
Replies
4
Views
877
Replies
4
Views
7K
  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
3
Views
937
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
7
Views
491
Top