Separable Differential Equations

I.  Homework

II.  Separable Differential Equations

A differential equation is called separable if it can be written as

f(y)dy = g(x)dx

To solve a separable differential equation

1)  Get all the y's on the left hand side of the equation and all of the x's on the right hand side.

2)  Integrate both sides.

3)  Plug in the given values to find the constant of integration (C)

4)  Solve for y

Example:

Solve dy/dx = y(3 - x);   y(0 )= 5

1)  dy/y = (3 - x)dx

2)  int dy/y = int (3 - x)dx

lny = 3x - x²

3)  ln5 = 0 + 0 + c

y = exp(3x - x² + ln5) = exp(3x - x²)exp(ln5)= 5exp(3x - x²)

Exercises:  

A)  dy/dx = x/y;   y(0) = 1

B)  dy/dx = x(x+1);  y(1) = 1

C)  2xy + dy/dx = x;  y(0) = 2

III.  Mixing

Suppose that you are a doctor and want to determine how long you should keep and IV flowing.  Assume that the IV lows at a rate of 2ml/min and has .1 grams of medication per ml.  Your patient has a total of 6000 ml of fluid in his system and excretes fluid at the same rate as the patient takes in.  How long should you let the IV run if you want to stop the IV when the patient has .01 gram per ml of medication in his system?

Solution:  We use the principle

Rate = Rate In- Rate Out

In particular, let x = number of grams of medicine in the body at time t.  Then the rate in is

Rate In = (2ml/min)(.1g/ml) = .2 g/min

and the rate out is

Rate Out = (2ml/min)(xg/6000ml) = x/3000 g/min

Hence

dx/dt = .2 - x/3000

Separating:

dx/(.2-x/3000) = dt

letting u = .2 - x/3000 du = -1/3000 dx gives

-3000 int du/u = int dt

-3000 ln|.2 - x/3000| = t + c

When t = 0 x = 0 hence

c = -3000ln(.2)

If the patient has .01 grams per ml and has 6000 ml of fluid, then the patient has 60g of medication. We solve

-3000 ln|.2 - 60/3000| = t - 3000 ln(.2)

t = -3000 ln|.2 - 60/3000| + 3000 ln(.2) = 316 minutes or about 5 hours and 16 minutes.