First order homogenous DE with variable coefficient

Click For Summary
SUMMARY

The discussion centers on solving the first order homogeneous differential equation represented as u' + (1/t)u = 0, derived from the second order equation x'' + (1/t)x' = 0 with initial conditions x(a) = 0 and x(1) = 1. The solution is established as x(t) = 1 - (log t)/(log a). Participants emphasized that the equation can be simplified by multiplying the second order ODE by t, leading to a more straightforward solution process without the need for reduction of order.

PREREQUISITES
  • Understanding of differential equations, specifically homogeneous types.
  • Familiarity with initial value problems in calculus.
  • Knowledge of logarithmic functions and their properties.
  • Experience with separation of variables technique in solving ODEs.
NEXT STEPS
  • Study the method of solving first order homogeneous differential equations.
  • Learn about the reduction of order technique in differential equations.
  • Explore the implications of initial conditions on the solutions of ODEs.
  • Investigate the application of logarithmic functions in differential equations.
USEFUL FOR

Students and professionals in mathematics, particularly those focusing on differential equations, as well as educators seeking to enhance their teaching methods in this area.

MrGandalf
Messages
30
Reaction score
0
I was stumped by this differential equation. The function x = x(t).

[tex]x^{\tiny\prime\prime} + \frac{1}{t}x^{\tiny\prime} = 0[/tex].

You have the initial values x(a) = 0 and x(1) = 1.

What I did was to introduce a new function u = x', so I ended up with the first order homogenous DE:
[tex]u^{\tiny\prime} + \frac{1}{t}u = 0[/tex].

This can't be that difficult, but the book which I am using does not give any details how to solve this particular problem. I only know how to solve it if I have a constant, or a function of t, on the right side.

The solution to the problem is:
[tex]x(t) = 1 - \frac{\log t}{\log a}[/tex].

Hope someone can shed some light on this. Thank ye, ye scurvy landlubber! Yarrr!
 
Last edited:
Physics news on Phys.org
When you did a reduction of order, you arrived to a separable equation that isn't difficult to solve at all. There is no need to reduce order since the original equation can be rewritten in a simpler form. What happens if you multiply the second order ode by t?. Can you simplify it (i.e. the derivative of something)?
 
Yes! Of course! How could I miss that? :)

Thank you for helping me!
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 3 ·
Replies
3
Views
7K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
1
Views
2K