Solving Integral Equations: Placing Arbitrary Constants in ln/e Solutions

  • Thread starter Thread starter Air
  • Start date Start date
  • Tags Tags
    Integration
Click For Summary

Homework Help Overview

The discussion revolves around solving a differential equation of the form y' = 3y, specifically focusing on the integration process and the placement of arbitrary constants in the solutions involving natural logarithms and exponentials.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to solve the differential equation and questions the placement of the arbitrary constant in their solution compared to the textbook's solution. Some participants point out a misunderstanding regarding the properties of exponentials.

Discussion Status

The discussion is ongoing, with participants exploring the correct application of exponential properties and clarifying the original poster's misunderstanding. There is acknowledgment of a mistake made by the original poster, but no consensus on the broader implications for the solution has been reached.

Contextual Notes

Participants are addressing the integration of the differential equation and the implications of arbitrary constants in solutions, with some confusion noted regarding the properties of exponentials.

Air
Messages
202
Reaction score
0
This is a question...

For the following question:
[itex]y^{'}=\frac{dy}{dx}=3y[/itex]

I get the solution...
[itex]\int \frac{1}{3y} dy = \int dx[/itex]
[itex]\frac{1}{3}ln y = x + c[/itex]
[itex]y = e^{3x}+e^{3c}[/itex]

However the textbook example says the solution is...
[itex]y = ce^{3x}[/itex]

My question is would my answer be incorrect? How should the arbitrary constant be placed in ln and e integral solutions?
 
Physics news on Phys.org
[itex]e^{x+y}=e^xe^y[/itex] and not [itex]e^{x+y}=e^x+e^y[/itex]. You made this mistake in your last line.
 
You have to take into account that
[tex]e^{x+y}=e^x \cdot e^y \neq e^x + e^y[/tex]
To get a feeling for that relation, take for example
[tex]2^{3+4}=(2 \cdot 2 \cdot 2 )\cdot (2 \cdot 2 \cdot 2 \cdot 2 )=2^3 \cdot 2^4[/tex]
 
Oh yes, that is correct. Silly mistake. Thanks.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 25 ·
Replies
25
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
4
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
3
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K