Solve Differential Equation: du/dt = e^(2u-16t) with Initial Condition u(0) = 0

[beanryu]

another question...

Solve the seperable differential equation for u
du/dt = e^(2u-16t)
Use the following initial condition: u(0) = 0

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.

here is what I did
du/dt = e^(2u)/e^(16t)
du/e^(2u) = dt/e^(16t)
ln(e^(2u))/2 = ln(e^(16t))/2+C
if u(0)=0... everything will be zero, including C

I know I am missing the hint
can someone point it out to me?! please...

one more last question
(2 pts) A tank contains 2300 liters of pure water. Beginning at time 0, solution containing 0.02 kg of sugar per liter enters the tank at a rate of 7 L/min. The solution is mixed and the mixed solution drains out of the tank at the same rate. Let y(t) be the amount of sugar (in kg) in the tank.
(a) Give the differential equation for y'(t) (in terms of the variable y)
all I know is that y=0.02*t
all i don't know how to y'(t) interms of y... enlighten me...

 

[AKG]

du/e^(2u) = dt/e^(16t)
ln(e^(2u))/2 = ln(e^(16t))/2+C

This is not right. It's true that:

$$\int \frac{du}{u} = ln(u)$$

but it's not true that:

$$\int \frac{du}{e^u} = ln(e^u)$$

or anything like it.

 

[beanryu]

gentlemens

I got the 1st question
but I need help on the second one
the tank one

please help
i got 1 hour to go... please THANKs in advance

 

[J77]

There was a mixing question on PF earlier in the week...