**Introduction to Difference Equations**

Q107. Our neighborhood association has a ten member board. Each year it plans to add four members. Write the difference equations that describe the size of the board (S) each year.

Year | Board Size |

1 | 10 |

2 | 14 |

3 | 18 |

4 | 22 |

S_{1} = 10

S_{2} = S_{1} + 4

S_{3} = S_{2} + 4 = (S_{1} + 4) + 4

S_{4} = S_{3} + 4 = ((S_{1} + 4) + 4) + 4

S_{4} = S_{1} + 4 + 4 + 4

S_{n} = S_{1} + (n - 1)4

p0=10

p1=p0+4

pn+1=pn+4

Ways systems behave: Exponential growth, linear growth, increase to a limit, oscillation, oscillate and go to a limit, and oscillate and grow.

Equilibrium is when: S_{n} = S_{n+1}

If you are graphing two lines, the point of intersection is the equilibrium.

Some systems move toward the equilibrium point (stable equilibrium) while others move away from it (unstable equilibrium).

Q108. You are a small non-profit. Your sole funder says that each year it will double what you have as your balance at the end of the year. Each year you project spending 20,000 for programs. Ignore interest. Write difference equations describing your balance (B).

Let's assume we start with balance B_{0}.

The first year we spend 20,000 and so at the end of the year have B_{0}−20,000 which the funder will double. Thus, our new balance will be

(1)

B_{1}=2×(B_{0}−20,000)

And, in general,

(2)

B_{n}+1=2×(B_{n}−20,000)

What special situations can you imagine we might get into?

Year | Balance | |

1 | (B_{0} - 20,000)*2 |
(B_{0} - 20,000)*2 |

2 | (B_{1} - 20,000)*2 |
(((B_{0} - 20,000)*2) - 20,000)*2 |

3 | (B_{2} - 20,000)*2 |
And so on…Each year can be |

4 | (B_{3} - 20,000)*2 |
expressed in terms of B_{0} |

Q109. Each year the feral cat population grows by 3%. Let Cn be the number of cats n years from now. Assume there are presently 350. Write a difference equation that describes the cat population from year to year.

C_{n} = 1.03^{n-1}*C_{1}

Year | C_{n} |

1 | C_{1} = 350 |

2 | C_{2} = 1.03*C_{1} = C_{1} + 0.03C_{1} |

3 | C_{3} = 1.03*C_{2} = 1.03(1.03*C_{1,)} |

n | Cn,, = 1.03^{n-1}*C_{1} |

What if the population decreases by 20 each year?

Year | C_{n} |

1 | C_{1} = 350 -20 |

2 | C_{2} = 1.03*C_{1} - 20 |

3 | C_{3} = 1.03*C_{2} - 20 = 1.03(1.03*C_{1} - 20) |

n | C_{n} = 1.03^{n-1}*C_{1} |

C_{n+1} = (1.03C_{n}) - 20

At equilibrium, C_{n} = C_{n+1} , therefore:

C_{n} = (1.03C_{n}) - 20

C_{n} - 1.03C_{n} = -20

-.03C_{n} = -20

.03C_{n} = 20

C_{n} = (20/.03) = 666.67

Year | Number of Cats |

1 | 350 |

2 | (350-20)1.03 = 339 |

3 | (339-20)1.03 = 329 |

4 | (329-20)1.03 = 318 |

5 | (318-20)1.03 = 307 |

6 | (307-20)1.03 = 295 |

7 | (395-20)1.03 = 284 |

8 | (318-20)1.03 = 272 |

9 | (272-20)1.03 = 259 |

10 | (259-20)1.03 = 246 |

Then graph it!

Hmm…we got different results than the equation above. Gotta check that out.