\( \frac { d y } { d t } = ( a - y ) ( b - y ) \)

The given differential equation is \[ \frac{dy}{dt} = (a - y)(b - y) \] where \( a \) and \( b \) are constants. This is a separable differential equation, and we can solve it by separating the variables \( y \) and \( t \). ### Step 1: Separate the variables We can rewrite the equation as: \[ \frac{dy}{(a - y)(b - y)} = dt \] ### Step 2: Integrate both sides Next, we need to integrate both sides. The left side requires partial fraction decomposition. We can express: \[ \frac{1}{(a - y)(b - y)} = \frac{A}{a - y} + \frac{B}{b - y} \] Multiplying through by the denominator \((a - y)(b - y)\) gives: \[ 1 = A(b - y) + B(a - y) \] Expanding this, we have: \[ 1 = Ab - Ay + Ba - By \] Combining like terms: \[ 1 = (A + B)(-y) + (Ab + Ba) \] Setting up the system of equations: 1. \( A + B = 0 \) 2. \( Ab + Ba = 1 \) From the first equation, we can express \( B = -A \). Substituting into the second equation gives: \[ Ab - A(a) = 1 \implies A(b - a) = 1 \implies A = \frac{1}{b - a} \] Thus, \[ B = -\frac{1}{b - a} \] Now we can rewrite the integral: \[ \int \left( \frac{1}{(a - y)} - \frac{1}{(b - y)} \right) dy = \int dt \] ### Step 3: Integrate Now we integrate both sides: \[ \int \frac{1}{(a - y)} dy - \int \frac{1}{(b - y)} dy = t + C \] The integrals yield: \[ -\ln|a - y| + \ln|b - y| = t + C \] ### Step 4: Simplify Combining the logarithms gives: \[ \ln\left|\frac{b - y}{a - y}\right| = t + C \] Exponentiating both sides results in: \[ \frac{b - y}{a - y} = e^{t + C} = Ke^t \] where \( K = e^C \) is a constant. ### Step 5: Solve for \( y \) Now we can solve for \( y \): \[ b - y = K e^t (a - y) \] Expanding and rearranging gives: \[ b - y = K a e^t - K y e^t \] \[ y + K y e^t = K a e^t - b \] Factoring out \( y \): \[ y(1 + K e^t) = K a e^t - b \] Thus, \[ y = \frac{K a e^t - b}{1 + K e^t} \] ### Conclusion The solution to the differential equation is: \[ y(t) = \frac{K a e^t - b}{1 + K e^t} \] where \( K \) is a constant determined by initial conditions.