inital conditions?

$f'(x)=\cfrac{x^2-1}{x}$: f(1)= 1/2, f(-1) = 0

Question

inital conditions?

$f'(x)=\cfrac{x^2-1}{x}$:  f(1)= 1/2, f(-1) = 0

angela210793 · Answer

u have to integrate this right?

amistre64 · Answer

yep

angela210793 · Answer

i mean in order to get f(x)

anonymous · Answer

I can do the integration Job 

$$f'(x) = x - \frac{1}{x}$$
$$f(x)=\frac{x^2}{2} + log_ex + C$$

amistre64 · Answer

good, now determine C by the inital conditions :)

anonymous · Answer

$$f(x)=\frac{x^2}{2} - log_ex + C$$

angela210793 · Answer

f(x)=x^2/2-lnx+c
then by f(1) and f(-1) u find c

anonymous · Answer

$$f(1) = \frac{1}{2} + C = \frac{1}{2}$$

anonymous · Answer

$$C = 0$$

amistre64 · Answer

i can get it for f(1) just fine ..., now try it with f(-1) = 0

amistre64 · Answer

when does ln|x| + C = 1/2 AND 0?

anonymous · Answer

Okay, Lol I see it now.

amistre64 · Answer

:)

amistre64 · Answer

this is a homework problem straight from the textbook ...

and i gotta wonder

angela210793 · Answer

Gosh...i hate everything tht includes integrals :/

amistre64 · Answer

C can be anything so I even tried:

ln(x) + ln(C) = ln(Cx)

ln(x) - ln(C) = ln(x/C)

zarkon · Answer

why are there two initial conditions?  are you sure you wrote the problem correctly?

zarkon · Answer

oh yea...

amistre64 · Answer

im sure .... im wondering if it aint spose to be a double derived

zarkon · Answer

so you have a +c1 for x<0 and a +c2 for x>0

zarkon · Answer

piecewise function

amistre64 · Answer

.... i gotta piece it in?  thats soo feels like cheating lol

anonymous · Answer

two different constants!... First time for me...

zarkon · Answer

$$f(x)=\left\{\begin{matrix}\frac{x^2}{2} -\ln|x| + C_1 & \text{ if } x>0 \ \frac{x^2}{2} - \ln|x|  + C_2 & \text{ if } x<0 \end{matrix}\right.$$

amistre64 · Answer

yep, the answer key for the one like it shows a piecewise as well ....

amistre64 · Answer

none of the examples in the chapter even give a hint to that ....

zarkon · Answer

this  "it is +c on evey open interval that is continiuous"

is probably buried somewhere in the text

zarkon · Answer

most of the time we ignore this

amistre64 · Answer

yeah, most of the time :)  thnx for the help yall

zarkon · Answer

sorry...not continuous...i mean defined

zarkon · Answer

your function is defined on (-infinity,0) and (0,infinity)

so we will have two constants

amistre64 · Answer

$$f(x)=\left\{\begin{matrix}\frac{x^2}{2} -\ln|x| & \text{ if } x>0 \ \frac{x^2}{2} - \ln|x| -\frac{1}{2} & \text{ if } x<0 \end{matrix}\right.$$looks good then