Initial Value Diff. Equation

1. Sep 29, 2007

robbondo

1. The problem statement, all variables and given/known data
Solve the Given initial value problem

2. Relevant equations

$$y' = tan ( x ), y ( pi / 4 ) = 3$$

3. The attempt at a solution

$$y = sec^2 ( x ) + 5 / 2$$

$$y = 3 - ln (sqrt(2) cos(x) )$$

Where did I got wrong? Or is the book screwed up?

2. Sep 29, 2007

bob1182006

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. Sep 29, 2007

HallsofIvy

Staff Emeritus
What do YOU believe is the integral of tan(x)??

4. Sep 29, 2007

Jim L

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. Sep 30, 2007

robbondo

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