I'm having trouble getting started. Steve said it's similar to another question I asked but I don't know how to begin. Any help?
Suppose a cell of volume V cc is surrounded by a homogeneous chemical solution of concentration C g/cc. Let y denote the concentration of the solute inside the cell at any time t and suppose that, initially, the concentration is y0. Fick's law, named after the German physiologist Adolf Fick (1829-1901), states that the rate of change of the concentration of solute inside the cell at any time t is proportional to the difference between the concentration of the solute outside the cell and the concentration inside the cell and inversely proportional to the volume of the cell. Fick's law may be expressed as the differential equation.
dy/dt = k/V(C-Y)
y(0) = y_0
Use separation of variables to solve the differential equation given above, where k, V, C, and y0 are constants with
C − y > 0.
(Note: The constant of proportionality k depends on the area and permeability of the cell membrane.)
y = ?????
check your previous posts. This one is explained there.
To solve the given differential equation using separation of variables, we need to separate the variables and integrate both sides of the equation.
The given differential equation is:
dy/dt = k/V(C - y)
To begin, let's rearrange the equation:
dy / (C - y) = (k / V) dt
Now, we can separate the variables by multiplying both sides by dt and dividing both sides by (C - y):
(1 / (C - y)) dy = (k / V) dt
Next, we integrate both sides of the equation. The integral of (1 / (C - y)) dy can be evaluated using the natural logarithm:
∫ (1 / (C - y)) dy = ∫ (k / V) dt
ln|C - y| = (k / V)t + C1
Here, C1 is the constant of integration.
Now, we can get rid of the absolute value by considering two cases:
Case 1: (C - y) > 0
In this case, ln|C - y| simplifies to ln(C - y):
ln(C - y) = (k / V)t + C1
Case 2: (C - y) < 0
In this case, we need to consider the absolute value:
ln|C - y| = -ln(-(C - y)) = ln(y - C)
Therefore, ln|C - y| simplifies to ln(y - C).
ln(y - C) = (k / V)t + C1
Now, let's solve for y.
In Case 1:
ln(C - y) = (k / V)t + C1
C - y = e^((k / V)t + C1)
y = C - e^((k / V)t + C1)
In Case 2:
ln(y - C) = (k / V)t + C1
y - C = e^((k / V)t + C1)
y = C + e^((k / V)t + C1)
Note: The values of C and C1 are determined by the initial condition given in the problem: y(0) = y0. You can substitute t = 0 and y = y0 into the equation to find the specific values of C and C1.
So, depending on the initial condition, the solution to the differential equation is either:
y = C - e^((k / V)t + C1)
or
y = C + e^((k / V)t + C1)