Query_16 

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course Mth 279

3/23 15:30

Question: Find the general solution toy '' - 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:

We can set up a characeristic Equation :

r^2 - 5 r + 2 = 0

r = 1 / 2 (5 ± sqrt(17) )

With this value of r we can set up the general solution

y = c1 e^( ½ (5 + sqrt(17)))t + c2 e^(1/2(5 - sqrt(17))) t

y(0) = - 1

y’(0) = -5

-1 = c1 + c2

-5 = c1(1/2(5 + sqrt(17)) + c2(1/2(5 - sqrt(17))

Solving simultaneously we have

c1 = -1.1

c2 = 0.1

The unique solution is

y = -1.1 e^( ½ (5 + sqrt(17))) + 0.1 e^(1/2(5 - sqrt(17)))

As t approaches infinity, the exponents become infitily large while the negative term taking precedence so the limit is negative infinity.

As t approaches negative infinity, the terms approach zero giving a limit of 0.

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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:

We can set up a characteristic equation of

8 r^2 - 6 r + 1 = 0

Using the quadratic formulas r equals

r = - 6 ± [sqrt(36 - 4(8)(1)) / 16]

r = ½ or ¼

The general solution is

y = c1 e^(t / 2) + c2 e^(t / 4)

y(1) = 4

y’(1) = 3 / 2

y(1) = c1 e^(1/2) + c2 e^(1/4) = 4

y’(1) = 1 / 2 c1 e^(1 / 2) + 1 / 4 c2 e^(1 / 4) = 3 / 2

Solving for c1 and c2 we have

c1 = 2 / sqrt(e)

c2 = 2 / 4throot(e)

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Given Solution:

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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:

m(r’’ - omega^2 r) = -k r’

m r’’ - m omega^2 r = - k r’

- m / k r’’ + m / k omega^2 r = r’

The general solution will be in the form

C1 e^(lambda t) + c2 e^(lambda t)

I am having trouble setting up the characteristic equation which would ostensibly supply two values for lambda, one positive and one negative.

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

If I divide out - m / k we have

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

Using the quadratic formula to solve for lambda we have

x = - b ± [sqrt( b^2 - 4 a c) / 2 a]

lambda = - k / m ± [sqrt( - k / m ^2 + 4 omega^2) / 2]

The value of lambda is getting ‘messy’

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You are on the right track.

You would have k^2 / m^2, not k / m^2, as the first term in that discriminant.

The roots would be

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

At this point I would recommend factoriing k/m out of the square root. You would get

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

You can then factor out k/m to get

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

This is a bit less messy, once you substitute the values of k / m and omega.

Your initial conditions will lead to an even less messy expression.

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I would then input this value of lambda into the general solution listed above, then use the initial conditions to solve for both c1 and c2.

??? Is my thought process correct so far?

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You are on the right track. Feel free to send me a copy of your solution when you get it.

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Given Solution:

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Self-critique (if necessary):

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