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

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. ### 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 \((a - y)(b - y)\) gives: \[ 1 = A(b - y) + B(a - y) \] Expanding this, we have: \[ 1 = Ab - Ay + Ba - By \] Combining like terms: \[ 1 = (Ab + Ba) - (A + B)y \] Setting coefficients equal, we get the system: 1. \( A + B = 0 \) 2. \( Ab + Ba = 1 \) From the first equation, we can express \( B \) in terms of \( A \): \[ B = -A \] Substituting into the second equation: \[ Ab - A(a) = 1 \implies A(b - a) = 1 \implies A = \frac{1}{b - a} \] Thus, \[ B = -\frac{1}{b - a} \] Now we can rewrite the integral: \[ \int \left( \frac{1}{b - a} \cdot \frac{1}{a - y} - \frac{1}{b - a} \cdot \frac{1}{b - y} \right) dy = \int dt \] ### Step 3: Integrate Integrating both sides gives: \[ \frac{1}{b - a} \left( -\ln |a - y| + \ln |b - y| \right) = t + C \] Combining the logarithms: \[ \frac{1}{b - a} \ln \left| \frac{b - y}{a - y} \right| = t + C \] ### Step 4: Solve for \( y \) Multiplying through by \( b - a \): \[ \ln \left| \frac{b - y}{a - y} \right| = (b - a)(t + C) \] Exponentiating both sides: \[ \left| \frac{b - y}{a - y} \right| = e^{(b - a)(t + C)} \] Let \( K = e^{(b - a)C} \), then: \[ \frac{b - y}{a - y} = K e^{(b - a)t} \] ### Step 5: Solve for \( y \) Cross-multiplying gives: \[ b - y = K e^{(b - a)t} (a - y) \] Rearranging: \[ b - y = K a e^{(b - a)t} - K y e^{(b - a)t} \] Combining terms involving \( y \): \[ y + K y e^{(b - a)t} = K a e^{(b - a)t} + b \] Factoring out \( y \): \[ y(1 + K e^{(b - a)t}) = K a e^{(b - a)t} + b \] Finally, solving for \( y \): \[ y = \frac{K a e^{(b - a)t} + b}{1 + K e^{(b - a)t}} \] This is the general solution to the differential equation. The constant \( K \) can be determined if an initial condition is provided.