# Another question

1. Apr 25, 2006

### 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 dont know how to y'(t) interms of y.... enlighten me....

Last edited: Apr 25, 2006
2. Apr 25, 2006

### AKG

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.

3. Apr 26, 2006

### beanryu

gentlemens

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