Question

Differential Equations - Cooling problem

Answer 1

http://www.math.wpi.edu/Course_Materials/MA1022A96/lab2/node5.html

Answer 2

let me know if this didn't help i can go through the question in steps

Answer 3

Yeah, I went through it, but I can't get the write answer

Answer 4

I know y initial = 100, and M = 20

Answer 5

so you start off with dy/dt = k(y-20)?

Answer 6

yep

Answer 7

and then make it seperable

so (1/y-20)dy = kdt

Answer 8

multiply both sides by

\[\frac{dt}{y-20}\] ... yep

Answer 9

dt/y-20 or dy/y-20?

Answer 10

so now you have \[\frac{dy}{y-20}=kdt\]

Answer 11

you got the seperation correct so really doesn't matter

Answer 12

now take the anti derivative of both sides

Answer 13

ln(y-20) = kt + c

Answer 14

yep

Answer 15

multiply by e

Answer 16

e^(ln(y-20)) = e^kt+c

Answer 17

y-20 = e^kt+c

Answer 18

\[e^{kt+c}=ce^{kt}\]

Answer 19

okay so now what do i do, this is the general formula right

Answer 20

yep so you take your initial values

Answer 21

The object was at 100 degrees at t=0

Answer 22

and t(5)=80

Answer 23

what does C equal? i got 80/e but i'm pretty sure that's wrong

Answer 24

if you put in the first initial you get
\[100-20=ce^0=c\]

Answer 25

80=c

Answer 26

now put in t=5 and solve

\[80-20=80e^{5k}\]

Answer 27

\[\frac{60}{80}=e^{5k}\]

Answer 28

8e^25 + 20?

Answer 29

are we solving for k now?

Answer 30

yes

Answer 31

okay, so you ln both sides?

Answer 32

yep you can simplify the left side too 3/4

Answer 33

\[ln(\frac{3}{4})=5k\]

Answer 34

divide by 5 and you get

\[k=\frac{ln(\frac{3}{4})}{5}\]

so now plug this into your original with c=80

\[y-20=80e^{\frac{ln(\frac{3}{4})t}{5}}\]

Answer 35

add 20 to both sides to get get

\[T=y=80e^{\frac{1}{5}ln(\frac{3}{4})t}+20\]

Answer 36

technically none are the answer because he's missing a t but i think that's just an error

Answer 37

Thanks, I got it! You helped a lot

Answer 38

no problem if you have any more DE questions i can probably help

Answer 39

That's it for now, but I'll make sure to contact you. Thanks