Commented
Posted a Comment: Sep 20th, 2011
"Where is your integrating factor and gen solution?"
Commented
Posted a Comment: Sep 20th, 2011
"I just got 30g. I think im going with this answer. happy to get you to confirm if you like."
Commented
Posted a Comment: Sep 20th, 2011
"0.297 i dont think can be correct. the solution gains mass. The intial value is 8grams. The maximum it can be is 200L x .4 = 80g"
Commented
Posted a Comment: Sep 20th, 2011
"hey, sorry for the late reply. i worked all day today. see how we go tonight. "
Commented
Posted a Comment: Sep 19th, 2011
"i found a mistake in mine. salt in is .4 x 20, not 4. x 10 as i had it written down. now im getting an 2/5 which is closer to their 3/4"
Commented
Posted a Comment: Sep 19th, 2011
"their solution looks ok, however im a little bit different
my powers cancelled out. did you get a specific solution?"
Commented
Posted a Comment: Sep 19th, 2011
"y(0) = 10g (100L x 0.1 grams) Intial condition,
so c = 50 (based on first post)
over flow will happen in 10 mins,
so y(10) = 30.25g
its late for me, i've got a few more days to double check, but seems about right. "simple" math says its going to gain 4grams a min (not countin"
Commented
Posted a Comment: Sep 19th, 2011
"or for v(t) = 100+10t
c(t) = 10+y/(100+10t)"
Commented
Posted a Comment: Sep 19th, 2011
"jhoijhoi, what did you get for your general solution for 3? i have (100t+10t)y(t)=20t(t+20) + C"
