Query 16

#$&*

course MTH 279

3/11 8 pm

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>? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: general: y= C1e^(6.6t) + C2e^(-1.6t) **approximations based on the value of 5/2 +- sqrt(17)

@& The characteristic equation is

r^2 - 5r + 2 =0

with solutions

r = 1/2(5 ± sqrt 17)

yielding solution set { e^((5 + sqrt(17) ) / 2 * t), e^((5 - sqrt(17) ) / 2 * t) }.

The 1/2 distributes over both terms.

*@

unique: y = -.80e^(6.6t) -.20e^(-1.6t) As t approaches infinity, the second part of the sum approaches 0, and the second part approaches infinity. As t approaches negative infinity, the first part of the sum approaches 0 and the second part approaches infinity.

@& According to your solution, the function would approach -infinity, not infinity, at both limits. Your reasoning on the limits is otherwise correct.

I believe the correct solution is

y = -1.1e^(1/2(5+ sqrt 17)t) + 0.1 *e^(1/2(5 - sqrt 17)t)

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. *@

confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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

Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK

********************************************* 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: general: y = C1e^(.5t) + C2e^(.25t) unique: 1.2e^(.5t) + 1.6e^(.25t) As t approaches infinity, the solution approaches infinity. As t approaches -infinity, the solution approaches 0.

@& Minor disagreement:

I get 1.6 for the second constant; I get 1.06 for the first.

Clearly you did the problem correctly, except for perhaps this one detail.

My system of equations is

c_1 e^(1/2) + c_2 e^(1/4) = 4 1/2 c_1 e^(1/2) + 1/4 c_2 e^(1/4) = 3/2

Check this out if you have time and let me know if you still believe your solution is correct. *@

confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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

Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK

********************************************* 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 ). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Very generally speaking: r = C1e^[ k/(2m) + sqrt( (k/m)^2 + omega^2) )*t] + C2e^[ k/(2m) - sqrt( (k/m)^2 + omega^2) )*t] general approximation using given values (after changing eveything into standard units): r= C1e^6.5t + C2e^-2.5t unique solution using approximations : r=v_0/9e^6.5t - v_0/9e^-2.5t r(2) = 491 m approximately

@& That would be a very long way from the centrifuge.

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 / (m) * ( 1/2 +- sqrt( 1 + 4 m^2 omega^2 / k^2) )

yielding general solution

r(t) = A e^( (-k / (m) ( 1/2 + sqrt( 1 + 4 m^2 omega^2 / k^2) ) * t ) + B e^( (-k / (m) ( 1/2 - sqrt( 1 + 4 m^2 omega^2 / k^2) ) * t ).

Following this through I get r between 7 cm and 8 cm.

*@

confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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

Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I feel okay about my work, but that r I got was huge, so I may've made a calculation error or two. ------------------------------------------------ Self-critique rating:3"

`gr51

Query 16

#$&*

course MTH 279

3/11 8 pm

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>?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

general: y= C1e^(6.6t) + C2e^(-1.6t)

**approximations based on the value of 5/2 +- sqrt(17)

@&

The characteristic equation is

r^2 - 5r + 2 =0

with solutions

r = 1/2(5 ± sqrt 17)

yielding solution set { e^((5 + sqrt(17) ) / 2 * t), e^((5 - sqrt(17) ) / 2 * t) }.

The 1/2 distributes over both terms.

*@

unique: y = -.80e^(6.6t) -.20e^(-1.6t)

As t approaches infinity, the second part of the sum approaches 0, and the second part approaches infinity. As t approaches negative infinity, the first part of the sum approaches 0 and the second part approaches infinity.

@&

According to your solution, the function would approach -infinity, not infinity, at both limits. Your reasoning on the limits is otherwise correct.

I believe the correct solution is

y = -1.1e^(1/2(5+ sqrt 17)t) + 0.1 *e^(1/2(5 - sqrt 17)t)

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.

*@

confidence rating #$&*:

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

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

Given Solution:

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

Self-critique (if necessary): OK

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

Self-critique rating: OK

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

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:

general: y = C1e^(.5t) + C2e^(.25t)

unique: 1.2e^(.5t) + 1.6e^(.25t)

As t approaches infinity, the solution approaches infinity. As t approaches -infinity, the solution approaches 0.

@&

Minor disagreement:

I get 1.6 for the second constant; I get 1.06 for the first.

Clearly you did the problem correctly, except for perhaps this one detail.

My system of equations is

c_1 e^(1/2) + c_2 e^(1/4) = 4

1/2 c_1 e^(1/2) + 1/4 c_2 e^(1/4) = 3/2

Check this out if you have time and let me know if you still believe your solution is correct.

*@

confidence rating #$&*:

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

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

Given Solution:

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

Self-critique (if necessary): OK

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

Self-critique rating: OK

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

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 ).

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Very generally speaking: r = C1e^[ k/(2m) + sqrt( (k/m)^2 + omega^2) )*t] + C2e^[ k/(2m) - sqrt( (k/m)^2 + omega^2) )*t]

general approximation using given values (after changing eveything into standard units): r= C1e^6.5t + C2e^-2.5t

unique solution using approximations : r=v_0/9e^6.5t - v_0/9e^-2.5t

r(2) = 491 m approximately

@&

That would be a very long way from the centrifuge.

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 / (m) * ( 1/2 +- sqrt( 1 + 4 m^2 omega^2 / k^2) )

yielding general solution

r(t) = A e^( (-k / (m) ( 1/2 + sqrt( 1 + 4 m^2 omega^2 / k^2) ) * t ) + B e^( (-k / (m) ( 1/2 - sqrt( 1 + 4 m^2 omega^2 / k^2) ) * t ).

Following this through I get r between 7 cm and 8 cm.

*@

confidence rating #$&*:

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

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

Given Solution:

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

Self-critique (if necessary): I feel okay about my work, but that r I got was huge, so I may've made a calculation error or two.

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

Self-critique rating:3"

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