Solving the Differential Equation dy/dt = y(y-1)(y-2) with Initial Condition y0 ≥ 0

This is a differential equation that describes the rate of change of a variable y with respect to time t. The equation states that the rate of change of y is equal to y times (y-1) times (y-2). The condition y0 ≥ 0 means that the initial value of y is non-negative.

To solve this differential equation, we can use separation of variables:

dy / y(y-1)(y-2) = dt

We can break down the fraction on the left-hand side using partial fraction decomposition:

1/y(y-1)(y-2) = A/y + B/(y-1) + C/(y-2)

Multiplying both sides by y(y-1)(y-2), we get:

1 = A(y-1)(y-2) + B(y)(y-2) + C(y)(y-1)

Setting y = 0, we get:

1 = 2A

So A = 1/2. Similarly, setting y = 1 and y = 2, we get:

1 = -B/2

1 = C/2

So B = -2 and C = 2. Therefore, we can rewrite the fraction as:

1/y(y-1)(y-2) = 1/2y - 1/(y-1) + 1/2(y-2)

Substituting this into the original differential equation, we get:

dy/dt = 1/2y - 1/(y-1) + 1/2(y-2)

This is now a separable differential equation that we can solve by integrating both sides:

∫dy / (y(y-1)(y-2)) = ∫dt

∫(1/2y - 1/(y-1) + 1/2(y-2)) dy = ∫dt

1/2 ln |y| - ln |y-1| + 1/2 ln |y-2| = t + C

Taking the exponential of both sides and simplifying, we get:

y(t) = (1 + 2e^{2t}) / (1 + e^{2t})

This is the general solution to the differential equation, which describes the behavior of the variable y over time. The condition y0 ≥ 0 means that we need to choose the constant C in such a way that y(0) is non-negative. We can solve for C using the initial condition:

y(0) = (1 + 2e^{0}) / (1 + e^{0}) = 3/2

So C = ln 3/2 - 1/2 ln 2. Therefore, the solution to the differential equation with initial condition y(0) = 3/2 is:

y(t) = (1 + 2e^{2t}) / (1 + e^{2t}) - ln 3/2 + 1/2 ln 2

This function describes the behavior of the variable y over time, where y represents a quantity that grows or decays according to the rate of change described by the differential equation.