Question

find f
\[f''(x) = x^{-2}, \quad x > 0,\quad f(1) = 0, \quad f(2) = 0\]

Answer 1

To find f
you need to first find f'
the find f' given f'' integrate both sides

Answer 2

hmm.. \[f'(x) = \frac 1x + c\]
??

Answer 3

almost

Answer 4

you are missing a certain constant multiple

Answer 5

ahh \[f'(x) = -\frac 1x + c\]

Answer 6

yes :)
now to find the constant
hmmm....you are missing a certain initial condition to do that ....

your question doesn't make sense ....
you need one of those to be f'(something)=another something

Answer 7

or maybe i should do it \[f'(x) = -\frac 1x + c_1\]

Answer 8

i might have typoed

Answer 9

integrate again

Answer 10

ah yes i did. it's f'(2) not just f(2)

Answer 11

get a system of 2 eau and 2 unknowns

Answer 12

hmm \[f(x) = -\ln x + c_1 x + c_2\]
yes?

Answer 13

ok great.
you can find that constant by using f(1)=0

and then do what zarkon says to find f

Answer 14

you can do the problem with two given values of f

Answer 15

or you can wait to find the first constant whatever

Answer 16

i suppose x > 0 is just there to note that -ln x exists?

Answer 17

oh wait....
i guess you can do it with f(something1)=another something1

and f(something2)=another something2

oops

Answer 18

f'(2) = 0

so f'(2) = -1/2 + c_1 = 0

does this mean c_1 is 1/2?

Answer 19

yes adding 1/2 to both sides solves that equation for c_1

Answer 20

then f(1) = -ln (1) + 1/2 x + c_2 = 0

so c_2 is -1/2?

Answer 21

x is 1
so you have -ln(1)+1/2(1)+c_2=0
and yes

Answer 22

oh. yeah...forgot to sub 1 into x the second time

Answer 23

so \[f(x) = -\ln x + \frac 12 x - \frac 12\]
??

Answer 24

tep

Answer 25

nice. thanks

Answer 26

i just noticed this was my 1000th question