Reduction of Order not giving a second solution

  • Thread starter Thread starter 1MileCrash
  • Start date Start date
  • Tags Tags
    Reduction
Click For Summary
SUMMARY

The discussion centers on solving the differential equation (x-1)y'' - xy' + y = 0, with a known solution y=e^x. The user initially assumes a second solution of the form ve^x, leading to an incorrect conclusion regarding the solution's validity. The error was identified as a failure to integrate v', which is crucial for finding the second solution. The correct approach involves re-evaluating the integration of v' and applying the appropriate transformations to derive the second solution.

PREREQUISITES
  • Understanding of second-order linear differential equations
  • Familiarity with the method of reduction of order
  • Knowledge of integrating factors and their application
  • Proficiency in manipulating exponential functions and logarithms
NEXT STEPS
  • Study the method of reduction of order in detail
  • Practice integrating functions involving exponential and logarithmic terms
  • Explore the application of integrating factors in differential equations
  • Review examples of second-order linear differential equations with known solutions
USEFUL FOR

Students studying differential equations, mathematics educators, and anyone seeking to deepen their understanding of solving second-order linear differential equations using reduction of order.

1MileCrash
Messages
1,338
Reaction score
41

Homework Statement



(x-1)y'' - xy' + y = 0, y=e^x is a solution

Homework Equations


The Attempt at a Solution



Assume the second solution is of the form ve^x, where v' = (y^'2)e^-int[-x/(x-1)]

So v' = e^(-2x)e^(x+ln|x-1|) = e^(ln|x+1|-x)

Then, this second solution must be

(e^x)(e^(ln|x-1|-x))

=e^(ln|x-1)

=x-1

But, this is no solution to the DE. What went wrong?

Thank you.
NEVERMIND, I forgot to integrate v'. What was I thinking?
 
Last edited:
Physics news on Phys.org
You mean: $$y^{\prime\prime} - \frac{x}{x-1}y^\prime + \frac{1}{x-1}y = 0$$ $$\frac{d}{dx}\left ( v^\prime(x)e^{2x}e^{\int \frac{xdx}{x-1}} \right )=0$$
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
2
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
7
Views
2K
Replies
3
Views
2K
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K