hi i need help with differentiation equation

- anonymous

hi i need help with differentiation equation

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- anonymous

post yur question

- anonymous

Solve the seperable differential equation for u

- anonymous

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

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## More answers

- anonymous

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

- anonymous

u*

- anonymous

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

- anonymous

can you solve from there?

- anonymous

this is calculus btw

- anonymous

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

- anonymous

i did do that wat u rewritten

- anonymous

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

- anonymous

is this calculus 1 ?

- anonymous

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

- anonymous

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

- anonymous

i found the constant to be -3/16

- anonymous

good?

- anonymous

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

- anonymous

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

- anonymous

is that constant correct?

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

wasnt mentioned in my question

- anonymous

i found c = 0

- anonymous

agree with MathMind.. c=0

- anonymous

c = 0
u = 4t

- anonymous

is that answer?

- anonymous

yup

- anonymous

it doesnt work, my homework dont accept that

- anonymous

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

- anonymous

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.

- anonymous

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

- anonymous

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

- anonymous

'oh wait nvm.. inik is right

- anonymous

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

- anonymous

can u figure out the u now?

- anonymous

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

- anonymous

so u = ?

- anonymous

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

- anonymous

I'm doing it now...

- anonymous

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

- anonymous

that is where i am confused at

- anonymous

give me sec

- anonymous

ok

- anonymous

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

- anonymous

it only takes away e^?

- anonymous

wat if it is -e^

- anonymous

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

- anonymous

now do i plug in 0 for t?

- anonymous

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)

- anonymous

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

- anonymous

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

- anonymous

not for general solution

- anonymous

what you mean?

- anonymous

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

- anonymous

oh haha alriught

- anonymous

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 :)

- anonymous

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

- anonymous

np :D thanks for joining

- anonymous

i have to go now baii

- anonymous

ty to u 2 mathmind
cya

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