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Question
Math
Posted 9 months ago

Which graph could be a solution to the logistic differential equation dNdt=5N(1N8)?\frac{d N}{d t}=5 N \cdot\left(1-\frac{N}{8}\right) ?

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Answer from Sia
Verified answer
Posted 9 months ago
Solution by Steps
step 2
This is a first-order nonlinear ordinary differential equation
step 3
To solve it, we use the method of separation of variables. Rewrite the equation as: dNN(1N8)=5dt\frac{dN}{N(1 - \frac{N}{8})} = 5 \, dt
step 4
Integrate both sides: dNN(1N8)=5dt\int \frac{dN}{N(1 - \frac{N}{8})} = \int 5 \, dt
step 5
The left side can be integrated using partial fractions: (8N+88N)dN=5dt\int \left(\frac{8}{N} + \frac{8}{8 - N}\right) dN = \int 5 \, dt
step 6
This gives: 8lnN8ln8N=5t+C8 \ln|N| - 8 \ln|8 - N| = 5t + C
step 7
Simplify and solve for NN: lnN8N=5t+C8\ln\left|\frac{N}{8 - N}\right| = \frac{5t + C}{8}
step 8
Exponentiate both sides: N8N=e5t+C8\frac{N}{8 - N} = e^{\frac{5t + C}{8}}
step 9
Let eC8=Ke^{\frac{C}{8}} = K, then: N8N=Ke5t8\frac{N}{8 - N} = K e^{\frac{5t}{8}}
step 10
Solve for NN: N=8Ke5t81+Ke5t8N = \frac{8K e^{\frac{5t}{8}}}{1 + K e^{\frac{5t}{8}}}
step 11
The general solution is: N(t)=8e5tC+e5tN(t) = \frac{8 e^{5t}}{C + e^{5t}} where C=8KC = \frac{8}{K}
step 12
Given the initial condition N(0)=1N(0) = 1, solve for CC: 1=8C+11 = \frac{8}{C + 1}
step 13
This gives C=7C = 7, so the specific solution is: N(t)=8e5t7+e5tN(t) = \frac{8 e^{5t}}{7 + e^{5t}}
step 14
The solution N(t)=8e5t7+e5tN(t) = \frac{8 e^{5t}}{7 + e^{5t}} approaches 8 as tt \to \infty
step 15
Therefore, the graph that shows NN starting at 1 and leveling off near 8 is the correct one
A
Key Concept
Logistic Growth
Explanation
The logistic differential equation models population growth that starts exponentially but levels off as it approaches a carrying capacity.

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