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erma4kov [3.2K]
3 years ago
9

This problem focuses on using Polymath, an ordinary differential equation (ODE) solver, and also a non-linear equation (NLE) sol

ver.
(a) There are initially 500 rabbits (x) and 200 foxes (y) on Farmer Oat's property. Use Polymath or MATLAB to plot the concentration of foxes and rabbits as a function of time for a period of up to 500 days. The predator-prey relationships are given by the following set of coupled ordinary differential equations:
{dx}/{dt}=k_{1}x-k_{2}x\cdot y
{dy}/{dt}=k_{3}x-k_{4}x\cdot y
Constant for growth of rabbits k1 = 0.02 day-1
Constant for death of rabbits k1 = 0.00004 / (dayx no. of foxes)
Constant for growth of foxes after eating rabbits k1 = 0.0004 / (dayx no. of rabbits)
Constant for death of foxes k1 = 0.04 day-1
What do your results look like for the case of k3 = 0.00004 / (day x no. of rabbits) and tfinal = 800 days? Also, plot the number of foxes versus the number of rabbits. Explain why the curves look the way they do. Vary the parameter k1, k2, k3, and k4. Discuss which parameter can or cannot be larger than others. Write a paragraph describing what you find.
(b) Use Polymath or MATLAB to solve the following set of nonlinear algebraic equations:
x^{3}y-4y^{2}+3x=1
6y^{2}-9xy=5
with initial guess of x = 2, y = 2.
Mathematics
1 answer:
dezoksy [38]3 years ago
4 0

Answer:

here

Step-by-step explanation:

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Use any of the methods to determine whether the series converges or diverges. Give reasons for your answer.
Aleks [24]

Answer:

It means \sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6} also converges.

Step-by-step explanation:

The actual Series is::

\sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6}

The method we are going to use is comparison method:

According to comparison method, we have:

\sum_{n=1}^{inf}a_n\ \ \ \ \ \ \ \ \sum_{n=1}^{inf}b_n

If series one converges, the second converges and if second diverges series, one diverges

Now Simplify the given series:

Taking"n^2"common from numerator and "n^6"from denominator.

=\frac{n^2[7-\frac{4}{n}+\frac{3}{n^2}]}{n^6[\frac{12}{n^6}+2]} \\\\=\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{n^4[\frac{12}{n^6}+2]}

\sum_{n=1}^{inf}a_n=\sum_{n=1}^{inf}\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\ \ \ \ \ \ \ \ \sum_{n=1}^{inf}b_n=\sum_{n=1}^{inf} \frac{1}{n^4}

Now:

\sum_{n=1}^{inf}a_n=\sum_{n=1}^{inf}\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\\ \\\lim_{n \to \infty} a_n = \lim_{n \to \infty}  \frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\\=\frac{7-\frac{4}{inf}+\frac{3}{inf}}{\frac{12}{inf}+2}\\\\=\frac{7}{2}

So a_n is finite, so it converges.

Similarly b_n converges according to p-test.

P-test:

General form:

\sum_{n=1}^{inf}\frac{1}{n^p}

if p>1 then series converges. In oue case we have:

\sum_{n=1}^{inf}b_n=\frac{1}{n^4}

p=4 >1, so b_n also converges.

According to comparison test if both series converges, the final series also converges.

It means \sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6} also converges.

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