Query 16

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course MTH 279

4:12 pm 4/1

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

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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:

r^2 - 5r + 2 =0

r = 1/2(5 ą sqrt 17)

y = c1*e^(1/2(5+ sqrt 17)t) + c2*e^(1/2(5- sqrt 17)t)

y(0)= -1 = c1+ c2

y'(0)= -5 = c1*(1/2(5+ sqrt 17) + c2 *(1/2(5- sqrt 17)

c1≈ -1.1067

c2≈ 0.1067

y = -1.1067*e^(1/2(5+ sqrt 17)t) + 0.1067*e^(1/2(5+ sqrt 17)t)

y --> inf as t --> inf

y--> 0 as t--> -inf

confidence rating #$&*:#8232; Given Solution:

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Self-critique (if necessary):

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Self-critique rating:

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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>?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

r^2-3/4*r+1= 0

r= 1/12(3ą sqrt 5)

y = c1*e^(1/12(3+ sqrt 5)t) +c2*e^(1/12(3- sqrt 5)t)

y(1)= 4 = 1.547c1+ 1.0657c2

y'(1)= 3/2 = 0.675c1 + 0.0678c2

c1 = 2.16

c2 = 0.618

y=2.16*e^(1/12(3+ sqrt 5)t) +0.618*e^(1/12(3- sqrt 5)t)

Here again, being exponential, y -->inf as t--> inf and y--> 0 as t--> -inf

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:

r''+ k/m r' - omega^2 r= 0

x= 1/2(-k/m ą sqrt ((k/m)^2+4omega^2) which are always real.

r = c1e^(1/2(-k/m + sqrt ((k/m)^2+4omega^2)t) + c2*e^(1/2(-k/m - sqrt ((k/m)^2+4omega^2)t)

r(0) = 0 --> c1+c2 = 0--> -c1= c2

r'(0)= v_0 --> 1/2(-k/m + sqrt ((k/m)^2+4omega^2)*c1 + 1/2(-k/m - sqrt ((k/m)^2+4omega^2)*c2 = v_0 --> c1= v_0/[(1/2(-k/m + sqrt ((k/m)^2+4omega^2) -1/2(-k/m - sqrt ((k/m)^2+4omega^2)] and c2 = -v_0/[(1/2(-k/m + sqrt ((k/m)^2+4omega^2) -1/2(-k/m - sqrt ((k/m)^2+4omega^2)]

y= v_0/[(1/2(-k/m + sqrt ((k/m)^2+4omega^2) -1/2(-k/m - sqrt ((k/m)^2+4omega^2)] *e^(1/2(-k/m ą sqrt ((k/m)^2+4omega^2)t) - v_0/[(1/2(-k/m + sqrt ((k/m)^2+4omega^2) -1/2(-k/m - sqrt ((k/m)^2+4omega^2)] *e^(1/2(-k/m ą sqrt ((k/m)^2+4omega^2)t)

If r2= 2*r1 then A_CS2 = 4* A_CS1 and V2 = 8*V1 so volume changes faster than cross sectional area meaning m increases faster than k and thus the drag is less and less since m is in the denominator and k/m --> 0. At 5 times r A_CS = 25 and V = 125 meaning 25k/125m = 1/5(m/k) this ratio goes to 0 as r goes to inf.

Substitution makes r:

r =0.00247e^(4.03t)-0.00247e^(-0.03t)

r(2) = 7.82 m

I'll have to review the problem in the text, but this appears to be a fairly straightforward application of the drag coefficient and centripetal acceleration. Particles are assumed to be geometrically similar, all with the same drag coefficient k / m in s^-1 leading to a single constant-coefficient equation; this would imply that the particles are also of uniform size.
In any case compare with my solution below:
Using trial solution e^(lambda t) our characteristic equation will be
lambda^2+ k/m lambda - omega^2 = 0
with solutions
lambda = (-k / (2 m) +- sqrt(k^2 / m^2 + 4 omega^2) / 2 ) = -k / (2m) * ( 1 +- sqrt( 1 + 4 m^2 omega^2 / k^2) )
yielding general solution
r(t) = A e^( (-k / (2 m) ( 1 + sqrt( 1 + 4 m^2 omega^2 / k^2) ) * t ) + B e^( (-k / (2 m) ( 1 - sqrt( 1 + 4 m^2 omega^2 / k^2) ) * t ).
Now given that k / m = 4 s^-1 and omega = 20 rev / min = 2 pi / 3 rad / sec we can put some numbers into our solution.
Note that 1 + 4 m^2 omega^2 / k^2 = 1 + 4 (m / k)^2 * omega^2 = 1 + 4 omega^2 / (k/m)^2 = 1 + 4 (2 pi / 3 s^-1)^2 / (4 s^-1)^2 = 1 + pi^2 / 9, so that sqrt( 1 + 4 m^2 omega^2 / k^2) = sqrt( 1 + pi^2 / 9) = 1.45, approx..
Our function becomes approximately
r(t) = A e^( ( -2 s^-1) ( 1 + 1.45) t) + B e^(-2 s^-1) ( 1 - 1.45) t) = A e^(-4.9 t) + B e^(.9 t)
with velocity function
v(t) = r ' (t) = -4.9 A e^(-4.9 t) + .9 B e^(.9 t).
r(0) = 0 and v(0) = r ' (0) = 1 cm/s so
A + B = 0
-4.9 A + .9 B = 1
This yields approximate solution A = -.17, B = .17, both in units of cm.
Thus
r(t) = -.17 e^(-4.9 t) + .17 e^(.9 t)
and
r(2) = 1.03 cm.

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary): That seems like an awfully hair general solution to that equation but r seems to make sense when I subbed in and it comes out right for v_0 = 1 cm/s too.

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start

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.

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

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

r''+ k/m r' - omega^2 r= 0

x= 1/2(-k/m ą sqrt ((k/m)^2+4omega^2) which are always real.

r = c1e^(1/2(-k/m + sqrt ((k/m)^2+4omega^2)t) + c2*e^(1/2(-k/m - sqrt ((k/m)^2+4omega^2)t)

r(0) = 0 --> c1+c2 = 0--> -c1= c2

Substitution makes r:

r =0.00247e^(4.03t)-0.00247e^(-0.03t)

r(2) = 7.82 m

I'll have to review the problem in the text, but this appears to be a fairly straightforward application of the drag coefficient and centripetal acceleration. Particles are assumed to be geometrically similar, all with the same drag coefficient k / m in s^-1 leading to a single constant-coefficient equation; this would imply that the particles are also of uniform size.
In any case compare with my solution below:
Using trial solution e^(lambda t) our characteristic equation will be
lambda^2+ k/m lambda - omega^2 = 0
with solutions
lambda = (-k / (2 m) +- sqrt(k^2 / m^2 + 4 omega^2) / 2 ) = -k / (2m) * ( 1 +- sqrt( 1 + 4 m^2 omega^2 / k^2) )
yielding general solution
r(t) = A e^( (-k / (2 m) ( 1 + sqrt( 1 + 4 m^2 omega^2 / k^2) ) * t ) + B e^( (-k / (2 m) ( 1 - sqrt( 1 + 4 m^2 omega^2 / k^2) ) * t ).
Now given that k / m = 4 s^-1 and omega = 20 rev / min = 2 pi / 3 rad / sec we can put some numbers into our solution.
Note that 1 + 4 m^2 omega^2 / k^2 = 1 + 4 (m / k)^2 * omega^2 = 1 + 4 omega^2 / (k/m)^2 = 1 + 4 (2 pi / 3 s^-1)^2 / (4 s^-1)^2 = 1 + pi^2 / 9, so that sqrt( 1 + 4 m^2 omega^2 / k^2) = sqrt( 1 + pi^2 / 9) = 1.45, approx..
Our function becomes approximately
r(t) = A e^( ( -2 s^-1) ( 1 + 1.45) t) + B e^(-2 s^-1) ( 1 - 1.45) t) = A e^(-4.9 t) + B e^(.9 t)
with velocity function
v(t) = r ' (t) = -4.9 A e^(-4.9 t) + .9 B e^(.9 t).
r(0) = 0 and v(0) = r ' (0) = 1 cm/s so
A + B = 0
-4.9 A + .9 B = 1
This yields approximate solution A = -.17, B = .17, both in units of cm.
Thus
r(t) = -.17 e^(-4.9 t) + .17 e^(.9 t)
and
r(2) = 1.03 cm.

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary): That seems like an awfully hair general solution to that equation but r seems to make sense when I subbed in and it comes out right for v_0 = 1 cm/s too.

------------------------------------------------

start

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

#*&!

*********************************************

Question: Solve the equation

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

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

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

r''+ k/m r' - omega^2 r= 0

x= 1/2(-k/m ą sqrt ((k/m)^2+4omega^2) which are always real.

r = c1e^(1/2(-k/m + sqrt ((k/m)^2+4omega^2)t) + c2*e^(1/2(-k/m - sqrt ((k/m)^2+4omega^2)t)

r(0) = 0 --> c1+c2 = 0--> -c1= c2

Substitution makes r:

r =0.00247e^(4.03t)-0.00247e^(-0.03t)

r(2) = 7.82 m

I'll have to review the problem in the text, but this appears to be a fairly straightforward application of the drag coefficient and centripetal acceleration. Particles are assumed to be geometrically similar, all with the same drag coefficient k / m in s^-1 leading to a single constant-coefficient equation; this would imply that the particles are also of uniform size.
In any case compare with my solution below:
Using trial solution e^(lambda t) our characteristic equation will be
lambda^2+ k/m lambda - omega^2 = 0
with solutions
lambda = (-k / (2 m) +- sqrt(k^2 / m^2 + 4 omega^2) / 2 ) = -k / (2m) * ( 1 +- sqrt( 1 + 4 m^2 omega^2 / k^2) )
yielding general solution
r(t) = A e^( (-k / (2 m) ( 1 + sqrt( 1 + 4 m^2 omega^2 / k^2) ) * t ) + B e^( (-k / (2 m) ( 1 - sqrt( 1 + 4 m^2 omega^2 / k^2) ) * t ).
Now given that k / m = 4 s^-1 and omega = 20 rev / min = 2 pi / 3 rad / sec we can put some numbers into our solution.
Note that 1 + 4 m^2 omega^2 / k^2 = 1 + 4 (m / k)^2 * omega^2 = 1 + 4 omega^2 / (k/m)^2 = 1 + 4 (2 pi / 3 s^-1)^2 / (4 s^-1)^2 = 1 + pi^2 / 9, so that sqrt( 1 + 4 m^2 omega^2 / k^2) = sqrt( 1 + pi^2 / 9) = 1.45, approx..
Our function becomes approximately
r(t) = A e^( ( -2 s^-1) ( 1 + 1.45) t) + B e^(-2 s^-1) ( 1 - 1.45) t) = A e^(-4.9 t) + B e^(.9 t)
with velocity function
v(t) = r ' (t) = -4.9 A e^(-4.9 t) + .9 B e^(.9 t).
r(0) = 0 and v(0) = r ' (0) = 1 cm/s so
A + B = 0
-4.9 A + .9 B = 1
This yields approximate solution A = -.17, B = .17, both in units of cm.
Thus
r(t) = -.17 e^(-4.9 t) + .17 e^(.9 t)
and
r(2) = 1.03 cm.

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary): That seems like an awfully hair general solution to that equation but r seems to make sense when I subbed in and it comes out right for v_0 = 1 cm/s too.

------------------------------------------------

start

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

#*&!#*&!

Query 16

@& I've inserted notes on at least one problem. You're doing fine, but check my notes.*@