Why doesn't this diff eq. have two solutions?

  • Context: Graduate 
  • Thread starter Thread starter gsingh2011
  • Start date Start date
  • Tags Tags
    Diff eq
Click For Summary
SUMMARY

The differential equation x' = 2x(x - 13) with the initial condition x(0) = 20 has a unique solution due to the constraints imposed by the initial value problem. After separating and integrating, the equation ln|(x - 13)/x| = 26t + C is derived. Exponentiating both sides leads to the expression (x - 13)/x = C'e^(26t), where C' must be determined by the initial condition, resulting in C' = 7/20. This confirms that only one solution exists that satisfies the given initial condition.

PREREQUISITES
  • Understanding of differential equations, specifically first-order separable equations.
  • Familiarity with natural logarithms and their properties.
  • Knowledge of initial value problems and their significance in determining unique solutions.
  • Basic skills in manipulating exponential functions and constants.
NEXT STEPS
  • Study the theory of existence and uniqueness theorems for differential equations.
  • Learn about the implications of initial conditions in solving differential equations.
  • Explore the use of computational tools like Wolfram Alpha for solving differential equations.
  • Investigate the behavior of solutions to nonlinear differential equations.
USEFUL FOR

Students and professionals in mathematics, particularly those focusing on differential equations, as well as educators seeking to clarify concepts related to initial value problems and solution uniqueness.

gsingh2011
Messages
115
Reaction score
1
Say you have the diff eq. x'=2x(x-13); x(0)=20. After separating and integrating we get,

ln|(x-13)/x|=26t+C

From here, we raise e to the power of each side to get rid of the natural log. Does this get rid of the absolute value signs? If so, why? If not, why is there only one solution (as dictated by wolfram alpha)?
 
Physics news on Phys.org
Wolfram alpha gives a solution. It does not necessarily mean there are not other solutions.

As for this particular problem, you seem to be forgetting the "initial value" part of the intial value problem. If you take the exponential of both sides you get
[tex]\frac{x- 13}{x}= e^{26t+ C}= e^Ce^{26t}= C'e^{26t}[/tex]
where [itex]C'= e^C[/itex]

Yes, C' can be either positive or negative but to satisfy x(0)= 20, we must have
[tex]\frac{20- 13}{20}= \frac{7}{20}=C'[/tex]
so there is only one solution.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 36 ·
2
Replies
36
Views
9K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K