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anonymous
 5 years ago
hi i need help with differentiation equation
anonymous
 5 years ago
hi i need help with differentiation equation

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Solve the seperable differential equation for u

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Use the following initial condition: u(0) = 0 U = ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0first lets rewrite the equation:\[e^{4u16t} = e^{4u}*e^{16t} = {e^{4u} \over e^{16t}} \]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0can you solve from there?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ops typo, the last one should be => e^(16t) no negative cause it is as a fraction form

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i did do that wat u rewritten

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[{du \over dt} = {e^{4u} \over e^{16t} } \]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0calculus 2 and i did do e^4u/e^16t how to proceed ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0e(4u)du=e(16t)dt solve it for each variable: (1/4)* e(4u)=(1/16)*e^(16t) + cont simplify

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i found the constant to be 3/16

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok. So, now all we do is separate 't' terms and 'u' tems\[{1\over e^{4u}} du= {1 \over e^{16t}}dt\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes but i found constant to be 3/16.... what to do next to find u?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is that constant correct?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0mathmind, what happens to (1/4) & (1/16) coefficients (you have to get it after integration)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thats wat u get when u substitute both u and t with 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0initial condition u(o)=0, what about condition for t?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wasnt mentioned in my question

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0agree with MathMind.. c=0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it doesnt work, my homework dont accept that

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(1/4)e^(4u)=(1/16)e^(16t) +C 4u=C*(16t) if u(0)=0, co C=0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0HINT: 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
 5 years ago
Best ResponseYou've already chosen the best response.0same thing.. c will be equal 0.. that's just a shortcut.. kinda

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you didn't tell this before... if u=0 & t=0 C=1/4 +1/16 = 3/16

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0'oh wait nvm.. inik is right

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so C is not zero? inik do u agree with mathmind?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0can u figure out the u now?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0comes from: (1/4)e^(4u)=(1/16)e^(16t) +C (1/4)*1= (1/16)*1 +C we did it!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0just put value of C=3/16 in: (1/4)e^(4u)=(1/16)e^(16t) +C simplify & take ln to both sides

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how do i take log of both sides am not good with logs

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that is where i am confused at

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you take ln to take away the e^\[\ln (e^x) = x\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it only takes away e^?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[{1 \over 4} e^{4u}={1 \over 16}e^{16t} +C\] 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
 5 years ago
Best ResponseYou've already chosen the best response.0now do i plug in 0 for t?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0let'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
 5 years ago
Best ResponseYou've already chosen the best response.0t= variable unless you been asked to put t=o, you can't

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0not for general solution

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it was not for you... sorry. you posted your response the same time as me...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I 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
 5 years ago
Best ResponseYou've already chosen the best response.0have to go. Thank you MathMind & jophil  was fun!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0np :D thanks for joining

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i have to go now baii

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ty to u 2 mathmind cya
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