Why doesn't my integration work out?

- anonymous

Why doesn't my integration work out?

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- schrodinger

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- dumbcow

because its lazy ...

- Luigi0210

I like Pi

- anonymous

I have this differential equation:
|dw:1370986567830:dw|
With the conditions when t=0, s=0 and v=30
I can't get a valid constant!! please help! <3

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## More answers

- anonymous

@dumbcow thank you, so insightful :)

- dumbcow

haha anyway what does g represent?
are these v(t) and s(t) functions ?

- dumbcow

ok well i assumed "g" is a constant and solved diff equ for v(s)
\[\large v(s) = \sqrt{15g-ke^{-2s/15}}\]
using initial values, s=0 v=30 i solved for constant of integration
\[k = 15g -900\]
hope that helps

- anonymous

sorry had to log straight after asking the question, g is gravity, v is velocity and s displacement - i separated the variables to get this
|dw:1370994064170:dw|
and need to solve for S x

- dumbcow

you can integrate left side using substitution
let
u = 15g - v^2
du = -2v dv
\[\rightarrow -\frac{15}{2} \int\limits \frac{du}{u} = \int\limits ds\]

- anonymous

ah yeah i tried that i got: \[-15/2 \int\limits_{}\frac{ -2v }{ 15g-v^2 }{} = \int\limits_{}ds{}\]

- anonymous

and then

- anonymous

\[-7.5 (\ln15g-v^2) = s+C\]

- anonymous

Using conditions s=0 v=30

- anonymous

\[-7.5 \ln(147-900) = C\]

- anonymous

not valid?

- Jhannybean

Oh...all along i was thinking "g" was another function :\ Oops!

- anonymous

no worries! i didnt explain it properly was short of time

- dumbcow

did you see my earlier post ... before you find constant you need to solve for v

- Jhannybean

@KateLovesPie are you in class right now or taking a test? Just wondering.

- anonymous

no lol neither

- anonymous

@dumbcow please could u show how u got to that?

- dumbcow

sure, just to clarify what is it you are trying to solve in the end?

- anonymous

i need an expression for S :)

- dumbcow

oh ok i was solving for V

- anonymous

ah i see ok yeah i was almost there with the ln above but it's not valid for C

- dumbcow

hmm i think i know where you went wrong...since we want expression for s or s(v)
you will include constant on right side NOT left side, it may affect the sign
\[-7.5\ln(15g-v^{2}) +C = s\]
then s=0 , v=30
\[C = 7.5\ln(15g - 900)\]
maybe?

- Jhannybean

Good luck with your problem!

- anonymous

thank you! thats looking better, the problem is within the ln function being negative as its ln(-753) :/

- dumbcow

oh shoot i didn't notice that...yeah thats no good

- anonymous

lol it's a tricky one!! the only way would be if it was ln(v^2-15g) which is what the answer says but i dont understand why this method isn't working out

- dumbcow

yeah lets see if we can arrange the diff equ differently
\[15v \frac{dv}{ds} = -(v^{2} -15g)\]
\[15 \int\limits \frac{v}{v^{2} -15g} dv = - \int\limits ds\]
\[7.5 \ln(v^{2} -15g)+C = -s\]
\[C = -7.5\ln(900-15g)\]
yes ? :)

- anonymous

yayy thank you!! seems like the other method should've worked as well which is really confusing! thanks so much for your help :)

- dumbcow

your welcome
reason 1st method didn't work wasn't because math was wrong but just cuz it made function undefined for certain values of "v"

- anonymous

Good work. Are you in an elementary differential equations class @KateLovesPie? This seems like an air resistance problem.

- anonymous

ah i see, i was sure i must have gone wrong somewhere, that makes more sense now :D

- anonymous

it is actually :) and i'm doing a level further maths, its part of the mechanics option, exam next week o.o

- anonymous

@oldrin.bataku do the signs normally switch with air resistance problems?

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