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\( \frac { d y } { d t } = y ( y - 1 ) ^ { 2 } ( y - 2 ) ^ { 3 } ( y - 3 ) ^ { 4 } \)

Ask by Macdonald Kelly. in the United States
Jan 25,2025

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The differential equation \( \frac{dy}{dt} = y (y - 1)^2 (y - 2)^3 (y - 3)^4 \) can be solved by separating variables and using partial fraction decomposition. After integrating, the general solution is expressed implicitly as: \[ A \ln |y| + B \ln |y - 1| - \frac{C}{y - 1} + D \ln |y - 2| - \frac{E}{y - 2} + \frac{F}{2(y - 2)^2} + \cdots = t + C \] This equation cannot be solved explicitly for \( y \) in terms of \( t \) using elementary functions. Therefore, the solution is typically left in its implicit form or analyzed qualitatively to understand the behavior of solutions.

Solution

To solve the differential equation \[ \frac{dy}{dt} = y (y - 1)^2 (y - 2)^3 (y - 3)^4, \] we can follow these steps: ### 1. **Separate Variables** First, separate the variables \( y \) and \( t \): \[ \frac{dy}{y (y - 1)^2 (y - 2)^3 (y - 3)^4} = dt \] ### 2. **Integrate Both Sides** Integrate both sides to find \( t \) as a function of \( y \): \[ \int \frac{1}{y (y - 1)^2 (y - 2)^3 (y - 3)^4} \, dy = \int dt = t + C \] where \( C \) is the constant of integration. ### 3. **Partial Fraction Decomposition** The integral on the left side is complex due to the multiple factors in the denominator. To tackle this, we can use **partial fraction decomposition**, expressing the integrand as a sum of simpler fractions: \[ \frac{1}{y (y - 1)^2 (y - 2)^3 (y - 3)^4} = \frac{A}{y} + \frac{B}{y - 1} + \frac{C}{(y - 1)^2} + \frac{D}{y - 2} + \frac{E}{(y - 2)^2} + \frac{F}{(y - 2)^3} + \frac{G}{y - 3} + \frac{H}{(y - 3)^2} + \frac{I}{(y - 3)^3} + \frac{J}{(y - 3)^4} \] Finding the coefficients \( A, B, C, \ldots, J \) involves solving a system of equations obtained by equating the coefficients of corresponding powers of \( y \) on both sides. This process can be tedious but is straightforward with systematic algebraic manipulation or using computational tools. ### 4. **Integrate Term by Term** Once the partial fractions are determined, integrate each term individually: \[ \int \left( \frac{A}{y} + \frac{B}{y - 1} + \frac{C}{(y - 1)^2} + \cdots + \frac{J}{(y - 3)^4} \right) dy = A \ln |y| + B \ln |y - 1| - \frac{C}{y - 1} + D \ln |y - 2| - \frac{E}{y - 2} + \frac{F}{2(y - 2)^2} + \cdots + \text{constant terms} + C \] Each integral follows standard forms, such as: - \( \int \frac{1}{y - a} \, dy = \ln |y - a| + C \) - \( \int \frac{1}{(y - a)^n} \, dy = \frac{1}{1 - n} \cdot \frac{1}{(y - a)^{n-1}}} + C \) for \( n \neq 1 \) ### 5. **Express the General Solution** After integrating, the general solution will be an implicit equation involving logarithmic and rational functions of \( y \): \[ A \ln |y| + B \ln |y - 1| - \frac{C}{y - 1} + D \ln |y - 2| - \frac{E}{y - 2} + \frac{F}{2(y - 2)^2} + \cdots = t + C \] Solving explicitly for \( y \) in terms of \( t \) may not be feasible due to the complexity of the equation. Therefore, the solution is often left in this implicit form or expressed using special functions if applicable. ### 6. **Qualitative Analysis (Optional)** Beyond finding an explicit solution, understanding the behavior of solutions can be insightful: - **Equilibrium Points:** \( y = 0, 1, 2, 3 \) are equilibrium solutions since \( dy/dt = 0 \) at these points. - **Stability:** - **\( y = 0 \):** Unstable (source). Solutions move away from 0 as \( t \) increases. - **\( y = 1 \):** Semi-stable. Solutions approach 1 from one side and move away on the other. - **\( y = 2 \):** Unstable (source). Solutions move away from 2 as \( t \) increases. - **\( y = 3 \):** Stable (sink). Solutions approach 3 as \( t \) increases. - **Behavior Between Equilibria:** - For \( y < 0 \): \( dy/dt > 0 \) (increasing) - For \( 0 < y < 1 \): \( dy/dt < 0 \) (decreasing) - For \( 1 < y < 2 \): \( dy/dt < 0 \) (decreasing) - For \( 2 < y < 3 \): \( dy/dt > 0 \) (increasing) - For \( y > 3 \): \( dy/dt > 0 \) (increasing) ### 7. **Conclusion** The differential equation is separable and can be solved implicitly using partial fraction decomposition. However, the explicit solution in terms of elementary functions is complex and typically expressed implicitly. For practical purposes, numerical methods or qualitative analysis are often employed to understand the behavior of solutions.

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Mind Expander

This equation showcases a specific form of a separable differential equation. To solve it, we can separate the variables by rewriting it as \( \frac{dy}{y(y - 1)^2(y - 2)^3(y - 3)^4} = dt \). Integrating both sides will help us find the general solution, which typically reveals the behavior of the function \( y(t) \) over time. Solving this differential equation can provide insight into population models or other real-world phenomena where the growth rate is dependent on the population size and specific thresholds, such as environmental carrying capacities or critical resource limits. Understanding these dynamics can be crucial in fields like ecology, epidemiology, and economics!

Related Questions

Multiple Choice Identify the choice that best completes the statement or answers the question. Find any points of discontinuity for the rational function. 1. \( y=\frac{(x-7)(x+2)(x-9)}{(x-5)(x-2)} \) a. \( x=-5, x=-2 \) b. \( x=5, x=2 \) c. \( x=-7, x=2, x=-9 \) d. \( x=7, x=-2, x=9 \) 2. \( y=\frac{(x+7)(x+4)(x+2)}{(x+5)(x-3)} \) a. \( x=-5, x=3 \) b. \( x=7, x=4, x=2 \) c. \( x=-7, x=-4, x=-2 \) d. \( x=5, x=-3 \) 3. \( y=\frac{x+4}{x^{2}+8 x+15} \) a. \( x=-5, x=-3 \) b. \( x=-4 \) c. \( x=-5, x=3 \) d. \( x=5, x=3 \) 4. \( y=\frac{x-3}{x^{2}+3 x-10} \) a. \( x=-5, x=2 \) b. \( x=5, x=-2 \) c. \( x=3 \) d. \( x \) \( =-5, x=-2 \) 6. What are the points of discontinuity? Are they all removable? \[ y=\frac{(x-4)}{x^{2}-13 x+36} \] a. \( x=-9, x=-4, x=8 \); yes b. \( x=1, x=8, x= \) -8; no c. \( x=9, x=4 \); no d. \( x=-9, x=-4 \); no 7. Describe the vertical asymptote(s) and hole(s) for the graph of \( y=\frac{(x-2)(x-5)}{(x-5)(x+2)} \). a. asymptote: \( x=2 \) and hole: \( x=-5 \) b. asymptotes: \( x=-2 \) and hole: \( x=-5 \) c. asymptote: \( x=-2 \) and hole: \( x=5 \) d. asymptote: \( x=-2 \) and hole: \( x=-2 \) a. \( x=-3, x=-8 \); no b. \( x=5, x=-7, x=1 \); no c. \( x=-5, x=7, x=-1 \); yes d. \( x=3, x=8 \); yes 8. Describe the vertical asymptote(s) and hole(s) for the graph of \( y=\frac{x+1}{x^{2}+6 x+5} \). a. asymptote: \( x=-1 \) and hole: \( x=-1 \) b. asymptote: \( x=-1 \) and hole: \( x=-1 \) c. asymptotes: \( x=-1,-1 \) and hole: \( x=-1 \) d. asymptote: -5 and hole: \( x=-1 \), 9. Find the horizontal asymptote of the graph of \( y=\frac{7 x^{6}+7 x+3}{9 x^{5}+7 x+3} \). a. \( y=0 \) b. \( y=\frac{7}{9} \) c. no horizontal asymptote d. \( y=\frac{6}{5} \)
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