Question

Will award medal
Find the solution to the differential equation
(dB/dt)+4B=10 B(1)=90

Answer 1

@myininaya

Answer 2

\[\frac{dB}{dt}+4B=10 \\ \frac{dB}{dt}=10-4B \\ \frac{dB}{10-4B}=dt \]
integrate both sides

Answer 3

if the left hand side looks ugly
use a substitution

Answer 4

i can prolly integrate it from then
i just was confused by the seperation of variables

Answer 5

Oh
well do you understand how I separated it?

Answer 6

oh yea ive done like 20 of these

Answer 7

oh ok
well let me know if I can help anymore
i will be around a little longer

Answer 8

so you get -4B=B(e^t)-10

Answer 9

what did you on both sides before doing anything else

Answer 10

like directly after the integration

Answer 11

oh you have ln|10-4b|=t+c ----> |10-4b|=e^(t+c)

Answer 12

you are missing something in front of the ln part

Answer 13

u=10-4b
du=-4 dB
so
-1/4 du=dB

Answer 14

so we should have
\[\frac{-1}{4} \ln|10-4b|=t+c\]

Answer 15

then you can multiply the -4 on both sides
then do the whole e^x and ln(x) are inverse thing

Answer 16

B=-350e^(-4t)-10

Answer 17

oops and wherever I had little b I meant big B :p

Answer 18

that shouldnt change much

Answer 19

i mean at least for my answer

Answer 20

lol no not really
you have some really big numbers for your B

Answer 21

oh you apply the condition

Answer 22

yea...

Answer 23

hmm...I will show you what I did

Answer 24

\[\frac{-1}{4} \ln|10-4B|=t+C \\ \ln|10-4B|=-4t+C \\ 10-4B=e^{-4t+C} \\ -4B=e^{-4t+C}-10 \\ B=\frac{-1}{4} e^{-4t +C}+\frac{10}{4} \\ B=\frac{-K}{4} e^{-4t}+\frac{5}{2} \\ B=K_1e^{-4 t}+\frac{5}{2} \\ \text{ so applying the } B(1)=90 \\ 90=K_1e^{-4}+\frac{5}{2} \\ 90-\frac{5}{2}=K_1e^{-4} \\ \frac{175}{2}=K_1e^{-4} \\ \frac{175}{2}e^4=K_1\]

Answer 25

like in the first step I could have written -4C but
that is still a constant
and that is also why I recalled -k/4 another constant value

Answer 26

it doesn't matter if we multiply the unknown constant by another constant it is still the unknown constant that we must find

Answer 27

the unknown constant is -362.5

Answer 28

that depends on what you used as the equation to find the unknown constant as

Answer 29

the thing is ending up with the same equation

Answer 30

90=b(-1/4)(e^4t)+5/2

Answer 31

\[B=K_1e^{-4t}+\frac{5}{2} \\ \text{ where } K_1=\frac{175}{2}e^4\]

Answer 32

is b your constant?

Answer 33

yea

Answer 34

\[90=\frac{-b}{4}e^{4}+\frac{5}{2} \\ -360=be^{4}+10 \\ -350=be^{4} \\ b=\frac{-350}{e^4}\]
ok so you would have
\[B=\frac{350}{4e^4}e^{-4t}+\frac{5}{2} \]
but I believe 350/4 reduced to 175/2 :)

Answer 35

same exact equation
just a a bit of different name for the constant we were finding

Answer 36

oops wait I think there was a negative missing in your equation
one sec

Answer 37

yea i tried it like you did and my webwork was wrong

Answer 38

90=b(-1/4)(e^4t)+5/2
this was
B=b(-1/4)e^(-4t)+5/2
right

Answer 39

the first way was right

Answer 40

but not the one with this newer constant

Answer 41

anyways it was suppose to be B=b(-1/4)e^(-4t)+5/2

Answer 42

so apply your B(1)=90

Answer 43

alright
ill give it a whirl

Answer 44

\[90=b(\frac{-1}{4})e^{-4}+\frac{5}{2} \\ -360=be^{-4}-10 \\ -350=be^{-4} \\ -350e^4=b\]

Answer 45

the e^4 was suppose to be on top not on bottom

Answer 46

\[B=b (\frac{-1}{4})e^{-4t}+\frac{5}{2} \\ \text{ where } b=-350e^4 \]

Answer 47

(-362.5e^4)/4e^(-4)+(5/2)

Answer 48

well almost

Answer 49

neg * neg =pos

Answer 50

and you are missing the t in the ^(-4) thing

Answer 51

and where does 362.5 come from?

Answer 52

continuing from what I had earlier
\[\frac{-1}{4} \ln|10-4B|=t+C \\ \ln|10-4B|=-4t+C \\ 10-4B=e^{-4t+C} \\ -4B=e^{-4t+C}-10 \\ B=\frac{-1}{4} e^{-4t +C}+\frac{10}{4} \\ B=\frac{-K}{4} e^{-4t}+\frac{5}{2} \\ B=K_1e^{-4 t}+\frac{5}{2} \\ \text{ so applying the } B(1)=90 \\ 90=K_1e^{-4}+\frac{5}{2} \\ 90-\frac{5}{2}=K_1e^{-4} \\ \frac{175}{2}=K_1e^{-4} \\ \frac{175}{2}e^4=K_1\]
\[B=K_1e^{-4t}+\frac{5}{2} \\ \text{ where } K_1=\frac{175}{2}e^{4} \\ \\ B=\frac{175}{2}e^4e^{-4t}+\frac{5}{2} \\ B=\frac{175}{2}e^{4-4t}+\frac{5}{2} \\ B=\frac{5}{2}(35e^{4-4t}+1)\]
This is the same result you will get using your equation and your constant ...
\[B=b (\frac{-1}{4})e^{-4t}+\frac{5}{2} \\ \text{ where } b=-350e^4 \\ B=-350e^4(\frac{-1}{4})e^{-4t}+\frac{5}{2} \\ B=\frac{350}{4}e^{4-4t}+\frac{5}{2} \\ B=\frac{175}{2}e^{4-4t}+\frac{5}{2} \\ B=\frac{5}{2}(35e^{4-4t}+1)\]

Answer 53

I know the algebra looks scary

Answer 54

You just have to take your time and work through it

Answer 55

Please tell me if you don't understand something up .
Or if you think I need to work through each step one at a time with you.

Answer 56

@pmkat14 If you are still there, I'm going to leave in a few. Just letting you know if you had any question.