Solve the differential equation dy dt = ky(1 − y) with initial condition y(0) = y0.

Answer 1

This is a standard example of separation of variables. First note that the zeros of the expression y(1-y) can give us constant solutions, depending on the value of y₀. If y=1 (constant), then dy/dt = 0 = ky(1-y), because 1-y=0 in this case. Similarly if y=0 (constant), then dy/dt = 0 = ky(1-y). The only way that either of these two constant solutions can work is if y₀=0 or y₀=1. If those solutions work, we're done. Otherwise, moving forward we will assume that y ≠ 1 and y≠0.

First we move the dt over to the right and divide both sides by the expression y(1-y) to obtain a new differential equation.

dy/(y(1-y)) = kdt

where we leave the initial condition y(0)=y₀ the same.

Now, you aren't supposed to have differentials like dy and dt without accompanying integral signs, so put in some integral signs:

∫dy/(y(1-y)) = ∫kdt

Now we have reduced this problem to finding some indefinite integrals. The integral on the right is easier:

∫kdt = kt + C

The integral on the left requires us to use partial fractions. The two factors in the denominator are y and (1-y) so we expect 1/(y(1-y)) to be some combination of the two simpler terms 1/y and 1/(1-y).

1/(y(1-y)) = A/y + B/(1-y).

Multiply through by the expression y(1-y) to obtain a new equation

1 = A(1-y) + B(y) = A + (B-A) y

Since the constant coefficient on the left side is 1 and the constant coefficient on the right is A, we have A = 1. Similarly the coefficient of y on the left side is 0 and the coefficient of y on the right side is B-A, so B-A = 0. This means that B=A = 1, so

1/(y(1-y)) = 1/y + 1/(1-y).

Now we can get back to our integrals:

∫dy/(y(1-y)) = ∫1/y dy + ∫ 1/(1-y) dy

These are both familiar integrals, so combining with our previous bit we have

ln |y| - ln |1-y| + C' = ln | y/(1-y) | + C'

where C' is to distinguish from the previous integral. Putting it all together we have

ln | y/(1-y) | = kt + C

(subtracting over the constant C' and relabeling the result C-C' as just the constant C).

Now use the exponential equation to get

| y/(1-y) | = e^{kt +C} = A_0 e^{kt}

where A_0 = e^C.

Now the constant A_0 is positive (because it is the output of the exponential function) so if we remove the absolute value bars from the left side we simply replace A_0 with ±A_0, which we can simply call the (nonzero, but possibly negative) constant A.

Now our equation is

y/(1-y) = A e^{kt}.

Multiply both sides by the expression 1-y to get

y = Ae^{kt} - y Ae^{kt}

Add the expression yAe^{kt} to both sides in order to isolate the variable y:

y + y Ae^{kt} = y(1 + Ae^{kt}) = Ae^{kt}

so we can divide through by the bit in parentheses to get

y = (Ae^{kt})/(1 + Ae^{kt})

This is the general solution, where A can be any real number. If y(0) = y₀ then we have

y₀ = A/(1+A)

so we can multiply both sides by 1+A to get

y₀ + Ay₀ = A

and then subtract over the Ay₀ term to get

y₀ = A - Ay₀ = A(1-y₀)

so that

A = y₀/(1-y₀)

This means that the solution to the initial value problem is

y = (Ae^{kt})/(1 + Ae^{kt})

where A = y₀/(1-y₀)