Independent functions as solutions

  • Context: Undergrad 
  • Thread starter Thread starter Sathish678
  • Start date Start date
  • Tags Tags
    Functions Independent
Click For Summary
SUMMARY

The discussion centers on the Cauchy equation represented as d/dr(r^2dR/dr) = 6R, where three functions—r^2, r^(-3), and (1/7)r^6—were proposed as solutions. However, it is established that (1/7)r^6 is not a valid solution because it does not satisfy the linearity and homogeneity of the equation. The calculations demonstrate that substituting (1/7)r^6 into the equation yields a discrepancy due to the missing factor of (1/7), confirming that only two independent solutions exist for this second-order linear differential equation.

PREREQUISITES
  • Understanding of second-order linear differential equations
  • Familiarity with the Cauchy equation
  • Knowledge of differentiation techniques
  • Basic concepts of linearity and homogeneity in differential equations
NEXT STEPS
  • Study the properties of second-order linear differential equations
  • Learn about the method of solving Cauchy equations
  • Explore the concept of linear independence of solutions
  • Investigate the implications of homogeneity in differential equations
USEFUL FOR

Mathematicians, physics students, and anyone studying differential equations who seeks to deepen their understanding of solution independence and the properties of linear differential equations.

Sathish678
Messages
1
Reaction score
1
TL;DR
I found three independent functions as solutions for this equation
d/dr(r^2dR/dr) = 6R (cauchy equation)
r^2 , r^(-3) , (1/7)r^6.
I found three independent functions as solutions for this equation
d/dr(r^2dR/dr) = 6R (cauchy equation)
r^2 , r^(-3) , (1/7)r^6.
But i read that a second order linear differential eqn has only two independent solutions.
Why this happened?
 
Physics news on Phys.org
\frac17 r^6 is not a solution. If it were, then as the equation is linear and homogenous r^6 must itself be a solution. It is not: <br /> \frac{d}{dr} \left(r^2 \frac{d}{dr}(r^6) \right) = \frac{d}{dr} \left( 6 r^7 \right) = 42r^6 \neq 6r^6.
 
  • Like
Likes   Reactions: Sathish678
In a bit mote detail, if R= \frac{1}{7}r^6 then \frac{dR}{dr}= \frac{6}{7}r^5. So r^2\frac{dR}{dr}= \frac{6}{7}r^7 and then \frac{d}{dr}\left(r^2\frac{dR}{dr}\right)= 6r^6. That is NOT equal to 6R because it is missing the "\frac{1}{7}".
 
Last edited:

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 0 ·
Replies
0
Views
3K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 12 ·
Replies
12
Views
2K