Initial Value Diff. Equation

  • Thread starter robbondo
  • Start date
  • #1
robbondo
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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?
 

Answers and Replies

  • #2
bob1182006
492
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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
 
  • #3
HallsofIvy
Science Advisor
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What do YOU believe is the integral of tan(x)??
 
  • #4
Jim L
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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
 
  • #5
robbondo
90
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sheesh... I differentiated instead of integrated. It's been a long summer. Thanks for your time guys, appreciate it.
 

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