Note that y(0) = .0008 ( 5000 ) = 4 ml. Hence setting t=0 in the preceding equation:

1. Example 2 A group of friends are out drinking. Joe realizes that they have no designated driver. He measures his blood alcohol content and finds it to be .08%. He decides to sip a light beer while his blood alcohol content falls. Joe has 5 liters of blood in his body which pass through his kidneys at 120 ml/min. Each minute the kidneys remove 1.2 ml of liquid from his blood including five times the square of the ml of alcohol passing through. The light beer replaces each ml of liquid removed by the kidneys with new liquid containing .0012 ml of alcohol. How long must Joe wait until his blood alcohol content falls to .02% which he thinks is low enough to drive home safely? Solution Let y(t) denote the number of ml of alcohol in Joe’s blood after t minutes. Blood enters his blood at (1.2)(.0012) = .00144 ml per min. Each ml of Joe’s blood contains y/5000 ml of alcohol. Hence alcohol leaves Joe’s blood at ml per minute. Thus the rate of change of alcohol in Joe’s blood is:

2. This is a separable differentiable equation. Hence we divide this equation by 1 - 2y2and integrate with respect to t: We use the method of partial fractions to integrate this rational function. Since we write Equate the numerators of these fractions with equal denominators: Let Let Then

3. Note that y(0) = .0008(5000) = 4 ml. Hence setting t=0 in the preceding equation: Joe’s blood has .02% blood alcohol content wheny(t)=.0002(5000)=1. At that time: The sun was rising when Joe drove home!