Question

hi i need help with differentiation equation

Answer 1

post yur question

Answer 2

Solve the seperable differential equation for u

Answer 3

du/dt =e^(4u-16t)

Answer 4

Use the following initial condition: u(0) = 0
U = ?

Answer 5

u*

Answer 6

first lets re-write the equation:\[e^{4u-16t} = e^{4u}*e^{-16t} = {e^{4u} \over e^{-16t}} \]

Answer 7

can you solve from there?

Answer 8

this is calculus btw

Answer 9

ops typo, the last one should be => e^(16t) no negative cause it is as a fraction form

Answer 10

i did do that wat u rewritten

Answer 11

\[{du \over dt} = {e^{4u} \over e^{16t} } \]

Answer 12

is this calculus 1 ?

Answer 13

calculus 2 and i did do e^4u/e^16t how to proceed ?

Answer 14

e(-4u)du=e(-16t)dt
solve it for each variable:
(-1/4)* e(-4u)=(-1/16)*e^(-16t) + cont
simplify

Answer 15

i found the constant to be -3/16

Answer 16

good?

Answer 17

ok. So, now all we do is separate 't' terms and 'u' tems\[{1\over e^{4u}} du= {1 \over e^{16t}}dt\]

Answer 18

yes but i found constant to be -3/16.... what to do next to find u?

Answer 19

is that constant correct?

Answer 20

mathmind, what happens to (-1/4) & (-1/16) coefficients (you have to get it after integration)

Answer 21

thats wat u get when u substitute both u and t with 0

Answer 22

initial condition u(o)=0, what about condition for t?

Answer 23

wasnt mentioned in my question

Answer 24

i found c = 0

Answer 25

agree with MathMind.. c=0

Answer 26

c = 0
u = 4t

Answer 27

is that answer?

Answer 28

yup

Answer 29

it doesnt work, my homework dont accept that

Answer 30

(-1/4)e^(-4u)=(-1/16)e^(-16t) +C
-4u=C*(-16t)
if u(0)=0, co C=0

Answer 31

HINT: To determine the constant of integration after you integrated both sides, DO NOT take natural logs, but rather just set u = 0 and t = 0 with u and t both still in the exponents. After determining the constant, then you need to take logs on both sides to solve for u.

This is what the hint that my questioned offered.

Answer 32

same thing.. c will be equal 0.. that's just a shortcut.. kinda

Answer 33

you didn't tell this before...
if u=0 & t=0
C=-1/4 +1/16 = -3/16

Answer 34

'oh wait nvm.. inik is right

Answer 35

so C is not zero? inik do u agree with mathmind?

Answer 36

can u figure out the u now?

Answer 37

comes from:
(-1/4)e^(-4u)=(-1/16)e^(-16t) +C
(-1/4)*1= (-1/16)*1 +C

we did it!

Answer 38

so u = ?

Answer 39

just put value of C=-3/16 in:
(-1/4)e^(-4u)=(-1/16)e^(-16t) +C
simplify & take ln to both sides

Answer 40

I'm doing it now...

Answer 41

how do i take log of both sides am not good with logs

Answer 42

that is where i am confused at

Answer 43

give me sec

Answer 44

ok

Answer 45

you take ln to take away the e^\[\ln (e^x) = x\]

Answer 46

it only takes away e^?

Answer 47

wat if it is -e^

Answer 48

\[{-1 \over 4} e^{-4u}={-1 \over 16}e^{-16t} +C\]
taking the Ln of everything so we can solve for u:\[{-1 \over 4}\ln (e^{-4u}) = {-1 \over 16} \ln(e^{-16t}) + \ln C\]\[{-1\over4} (-4u) = {-1 \over 16} (-16t) + \ln C\]

Answer 49

now do i plug in 0 for t?

Answer 50

let's try:
e^(-4u)=1/4 * e^(-16t) +3/4
-4u=ln[1/4*(e^(-16t) +3)]
u=-1/4 *[ln(e^(-16t)+3) - ln4]
u=ln4/4 -1/4* ln(e^(-16t) +3)

Answer 51

t= variable
unless you been asked to put t=o, you can't

Answer 52

u(t)=...
u(0) = 0

Answer 53

not for general solution

Answer 54

what you mean?

Answer 55

it was not for you... sorry. you posted your response the same time as me...

Answer 56

oh haha alriught

Answer 57

I mean that the answer should be in form u(t)=...one thing
if u(o) means t=0 - another.
your response would be correct :)

Answer 58

have to go.
Thank you MathMind & jophil - was fun!

Answer 59

np :D thanks for joining

Answer 60

i have to go now baii

Answer 61

ty to u 2 mathmind
cya