# Initial Value Diff. Equation

robbondo

## Homework Statement

Solve the Given initial value problem

## Homework Equations

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

## 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?

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

Homework Helper
What do YOU believe is the integral of tan(x)??

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

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