Troubleshooting Initial Value Problems for Differential Equations

  • Thread starter Thread starter robbondo
  • Start date Start date
  • Tags Tags
    Initial Value
Click For Summary

Homework Help Overview

The discussion revolves around an initial value problem for a differential equation, specifically involving the equation y' = tan(x) with the initial condition y(π/4) = 3. Participants are examining the integration process and comparing their results with a textbook answer.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are discussing the integration of tan(x) and questioning the correctness of their own and the textbook's answers. There are attempts to clarify the integration process and the implications of the initial condition.

Discussion Status

Some participants have provided insights into potential errors in integration and have suggested alternative interpretations of the integral of tan(x). There is acknowledgment of differing results, but no explicit consensus has been reached regarding the correct solution.

Contextual Notes

Participants are navigating through the integration of trigonometric functions and the implications of initial conditions. There is a mention of a possible error in differentiation versus integration, which may affect the understanding of the problem.

robbondo
Messages
90
Reaction score
0

Homework Statement


Solve the Given initial value problem

Homework Equations



[tex]y' = tan ( x ), y ( pi / 4 ) = 3[/tex]

The Attempt at a Solution



Well I got the answer

[tex]y = sec^2 ( x ) + 5 / 2[/tex]

The books answer is

[tex]y = 3 - ln (sqrt(2) cos(x) )[/tex]

Where did I got wrong? Or is the book screwed up?
 
Physics news on Phys.org
you integrated wrong, i don't know how they got sqrt(2) but you need to change tan(x) to sin(x)/cos(x) and do a substitution to get the log
 
What do YOU believe is the integral of tan(x)??
 
I am no expert, but I get the books answer, based on y=- ln (cos x)+C, and C =3 - ln[(sqrt2) /2)] , which = 3+ ln (sqrt 2).

Or y=-ln (cos x) -ln ( sqrt 2 ) +3

y= - ln [ (cos x) sqrt 2] +3

Dont bet your life on it. Jim
 
sheesh... I differentiated instead of integrated. It's been a long summer. Thanks for your time guys, appreciate it.
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 40 ·
2
Replies
40
Views
5K
  • · Replies 11 ·
Replies
11
Views
3K
Replies
6
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
24
Views
3K