lgbasallote
find f
\[f''(x) = x^{-2}, \quad x > 0,\quad f(1) = 0, \quad f(2) = 0\]
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myininaya
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To find f
you need to first find f'
the find f' given f'' integrate both sides
lgbasallote
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hmm.. \[f'(x) = \frac 1x + c\]
??
myininaya
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almost
myininaya
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you are missing a certain constant multiple
lgbasallote
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ahh \[f'(x) = -\frac 1x + c\]
myininaya
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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
lgbasallote
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or maybe i should do it \[f'(x) = -\frac 1x + c_1\]
lgbasallote
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i might have typoed
Zarkon
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integrate again
lgbasallote
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ah yes i did. it's f'(2) not just f(2)
Zarkon
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get a system of 2 eau and 2 unknowns
lgbasallote
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hmm \[f(x) = -\ln x + c_1 x + c_2\]
yes?
myininaya
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ok great.
you can find that constant by using f(1)=0
and then do what zarkon says to find f
Zarkon
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you can do the problem with two given values of f
myininaya
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or you can wait to find the first constant whatever
lgbasallote
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i suppose x > 0 is just there to note that -ln x exists?
myininaya
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oh wait....
i guess you can do it with f(something1)=another something1
and f(something2)=another something2
oops
lgbasallote
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f'(2) = 0
so f'(2) = -1/2 + c_1 = 0
does this mean c_1 is 1/2?
myininaya
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yes adding 1/2 to both sides solves that equation for c_1
lgbasallote
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then f(1) = -ln (1) + 1/2 x + c_2 = 0
so c_2 is -1/2?
myininaya
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x is 1
so you have -ln(1)+1/2(1)+c_2=0
and yes
lgbasallote
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oh. yeah...forgot to sub 1 into x the second time
lgbasallote
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so \[f(x) = -\ln x + \frac 12 x - \frac 12\]
??
myininaya
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tep
lgbasallote
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nice. thanks
lgbasallote
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i just noticed this was my 1000th question