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MAT275, HW 1

DR. NEAL BUSHAW

All problems are from Boyce/DiPrima, 10th edition. Due date is Sept 11, 2013.

Assigned on 26 Aug

1.2.3. Consider the following differential equation:

dy/dt= ay+b,

where both a and bare positive numbers.

a) Find the general solution of the differential equation.

b) Sketch the solution for several different initial conditions.c) Describe how the solutions change under each of the following conditions:

i) a increases.ii) b increases.

Solution: a):dy/dt

yb/a=a

d/dt[ln|yb/a|] =a d/dt[ln|yb/a|]dt=

a dt

ln|yb/a|= at+c

yb/a= ceat

y= ceat +b/a

b): Just expecting a few graphs, with different values for c. These dont have to be verygood - since a, b are arbitrary, just looking for something that looks like an exponential.c):i): Again, this is a vague question, so something like as b increases, the asymptoteincreases.c):ii): as a increases, the asymptote decreases towards 0 and the rate of the exponentialincreases / exponential steepens / is stretched vertically /etc.

1.2.4. Consider the differential equation dy/dt= ay b.

a) Find the equilibrium solutionye.

b) Let Y(t) = y ye; thus Y(t) is the deviation from the equilibrium solution. Findthe differential equation satisfied by Y(t).

Date: 26 Aug.

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2 DR. NEAL BUSHAW

Solution: a):dy/dt= 0

ayb= 0

ay= b

y= b/a

. b): People may have done this MANY different ways; the way I was hoping for issomething like:

Y(t) =y ye

Y =y ye

Y =ayb0

Y =a(yb/a)

Y =aY.

1.2.7. The field mouse population in Example 1 satisfies the differential equation

dp/dt= 0.5p450.a) Find the time at which the population becomes extinct ifp(0) = 850.b) Find the time of extinction ifp(0) =p0, where 0< p0 < 900.c) Find the initial populationp0 if the population is to become extinct in 1 year.

Solution: a): dp/dt= 0.5p450, so they should get the general solution P(t) =ce0.5t +900. Then IVP: 850 = ce0.50 + 900 gives c = 50. Then, want extinction time:

0 =50e0.5t + 900

900 = 50e0.5t

18 =e0.5t

ln18 = 0.5t

2 ln 18 =t5.78 months= t

Missing that this is months here is okay; theyll get dinged in the third part for thatmistake. b): IVP:p0= c+ 900, so p0900 =c. Then same algebra as last time:

0 = (p0900)e0.5t + 900

900

p0900=e0.5t

ln 900

p0900= 0.5t

2 ln 900

p0900

=t

Any equivalent form to this is fine - if they used log rules to simplify the fraction, thatsgreat, if not, that is fine too. c):

P(12) = 0

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