Assignment 18

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course Mth 279

7/15

Query 16 Differential Equations*********************************************

Question: Find the general solution to

y '' - 5 y ' + 2 y = 0

and the unique solution for the initial conditions y (0) = -1, y ' (0) = -5.

How does the solution behave as t -> infinity, and as t -> -infinity>?

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Your solution:

1) y = Ae^(rt) ; y’ = rAe^(rt) y’’ = r^2Ae^(rt)

2) sub in and divide by Ae^(rt)

r^2 -5r + 2 = 0

3) use quadratic formula

r = -(-5) +- sqrt((-5)^2 - 4(1)(2))/2

r = 5+sqrt(17)/2 r= 5-sqrt(17)/2

4) General soln: y(t) = Ae^(4.561t) + Be^(.438t)

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Approximations of the coefficients of the exponential are inappropriate here.

Your general solution is

y = A e^(1/2(5+ sqrt 17)t) + B *e^(1/2(5 - sqrt 17)t)

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Find A and B

5) take derivative

y’ = 4.561Ae^(4.561t) + .438Be^(.438t)

6) plug in condition

y(0) = -1

-1 = Ae^0 + Be^0

y’(0) = -5

-5 = 4.561Ae^0 + .438Be^0

-1 = A + B

-5 = 4.561A + .438B

7) solve system of equations

B = -1 - A; sub into second equation

-5 = 4.561A + .438(-1-A)

-5 = 4.561A -.438 -.438A

-4.562 = 4.128A

A = -1.105

-1 = -1.105 + B

B = 0.105

unique soln:

y(t) = -1.105e^4.561t + 0.105e^.438t

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Approximations of the values of A and B are OK, and yours are correct.

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how soon behaves?

t -> infinity

plug in pos values to see pattern of function

as t->infinity y -> infinity

as t-> -infinity y->0

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As t -> infinity the exponents of both both terms remain positive and become large, with the negative term having the greater magnitude; as a result the limit is -infinity.

As t -> - infinity both exponents approach -infinity so both terms approach zero, resulting in a limit of zero.

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Unsure of the interval question???

confidence rating #$&*:

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Given Solution:

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Question: Find the general solution to

8 y '' - 6 y ' + y = 0

and the unique solution for the initial conditions y (1) = 4, y ' (1) = 3/2.

How does the solution behave as t -> infinity, and as t -> -infinity>?

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Your solution:

y = Ae^(rt)

y’= rAe^(rt)

y’’= r^2Ae^(rt)

8r^2 - 6r + 1 = 0

use quad ration formula

r= -(-6) +- sqrt((-6)^2 - 4(8))/(2*8)

r = 6+- sqrt(36-32)/16

r = 1/2 and r = 1/4

general soln: y(t) = Ae^(.5t) + Be^(.25t)

take derivative

y’ = .5Ae^(.5t) + .25Be^(.25t)

y(1) = 4

4 = Ae^.5 + Be^.25

y’(1) = 3/2

3/2 = .5Ae^.5 + .25Be^.25

4 = Ae^.5 + Be^.25

3/2 = .5Ae^.5 + .25Be^.25 (multiply by 4 then subtract)

4 = Ae^.5 + Be^.25

-6 = 2Ae^.5 +Be^.25

-2 = -1Ae^.5

A=2e^-.5 = 1.21

4= (2e^-.5)(e^.5) + Be^.25

4 = 2 + Be^.25

B = 2e^-.25 = 1.55

unique soln:

y(t) = 1.21e^(.5t) +1.55e^(.25t)

at t->infinity y increases y-> infinity

as t-> -infinity y -> 0

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

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Self-critique rating:

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Question: Solve the equation

m ( r '' - Omega^2 r) = - k r ' for r(0) = 0, r ' (0) = v_0.

Omega is the angular velocity of a centrifuge, m is the mass of a particle and k is drag force constant. Physics students will recognize that m r Omega^2 is the centripetal force required to keep an object of mass m moving in a circle of radius r at angular velocity Omega.

The equation models the motion of a particle at the axis which is given initial radial velocity v_0.

The mass m of a particle is proportional to its volume, while the drag constant is proportional to its cross-sectional area. Assuming all particles are geometrically similar (and most likely spherical, though this is not necessary as long as they are geometrically similar, but for the sake of an accurate experiment spheres would be preferable), how then does k / m change as the diameter of particles increases?

If Omega = 20 revolutions / minute and v_0 = 1 cm / second, with k / m = 4 s^-1, find r( 2 seconds ).

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Your solution:

mr’’ - m(omega^2)r = -kr’

r’’ + k/m r’ - omega^2 r = 0

x^2 + k/m x - omega^2 = 0

solve for x

x = -k/m +- sqrt(k^2/m^2 - 4(-omega^2))/2

general form

r(t) = Ae^((-k/m +- sqrt(k^2/m^2 - 4(-omega^2))/2)t + Be^((-k/m +- sqrt(k^2/m^2 - 4(-omega^2))/2)t)

plug in some values

k/m = 4

omega = 20rev/min(1min/60s)(2pi) = 2pi/3

r(t) = Ae^((-4+sqrt(96)/2)t) + Be^((-4-sqrt(96)/2)t)

r(0)=0

0 = A + B ; A = -B

r’(0) = 1

r’(t) = 2.89A + (-6.89)B = 1

2.89*-B - 6.89B = 1

-9.78B = 1

B = -0.102

A = 0.102

r(t) = 0.102e^(2.89t) - 0.102e^-6.89t

So r(2) = 0.102e^(2.89*2) - 0.102 e^(-6.89*2)

r(2) = 33.02cm

&#Your work looks good. See my notes. Let me know if you have any questions. &#