Query 16

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course MTH 279

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:

We look for solutions in the form y(t) = e^(λt).

d/dt(e^(λt)) = λe^(λt); d^2/dt^2(e^(λt)) = λ^2e^(λt);

Substituting this back into the original DE gives:

λ^2*e^(λt) - 5λe^(λt) + 2e^(λt) = 0.

Factoring yields:

e^(λt)[λ^2 - 5λ + 2] = 0

λ^2 - 5λ + 2 = 0, which is the characteristic polynomial for the differential equation.

Solving for the roots, or zeros, of the polynomial, using the quadratic formula, we get:

λ = (5±sqrt((-5)^2 - 4*2))/2 = (5±sqrt(17))/2 ≈ 0.43845, 4.5616;

y_1(t) = e^((5+sqrt(17))/2*t); y_2(t) = e^((5-sqrt(17))/2*t);

y_g(t) = c_1*e^((5+sqrt(17))/2*t) + c_2e^((5-sqrt(17))/2*t) is the general solution.

y_g’(t) = ((5+sqrt(17))/2)c_1e^((5+sqrt(17))/2*t) + ((5-sqrt(17))/2)c_2e^((5-sqrt(17))/2*t);

Imposing the initial conditions (y(0) = -1, y’(0) = -5) and solving simultaneously, we get:

y_g(0) = c_1 + c_2 = -1

y_g’(0) = ((5+sqrt(17))/2)c_1 + ((5-sqrt(17))/2)c_2 = -5

c_1 = (-5sqrt(17))/34 - ½ ≈ -1.10634

c_2 = (5sqrt(17))/34 - ½ ≈ 0.10634

y(t) = [-(5sqrt(17))/34 - ½]e^((5+sqrt(17))/2*t) + [(5sqrt(17))/34 - ½]e^((5-sqrt(17))/2*t);

Taking the limits at the appropriate bounds:

lim(y(t),t,-∞) = 0-0 = 0; lim(y(t),t,∞) = -∞

As t->-∞, the solution approaches 0. As t->∞, the solution approaches -∞.

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary): OK

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Self-critique rating: OK

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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’’-3/4y’+1/8y=0

Assume that solutions will take the form of y(t) = e^(λt).

y’(t) = λe^(λt)

y”(t) = λ^2*e^(λt)

Substituting back in:

λ^2*e^(λt) - 3λe^(λt)/4 + e^(λt)/8 = 0

e^(λt)(λ^2-3/4λ+1/8) = 0

λ^2-3/4λ+1/8 = 0

Quadratic formula yields:

λ=1/4,1/2

y_1 = e^(t/2); y_2 = e^(t/4)

y_g = c_1e^(t/2) + c_2e^(t/4)

y_g’ = c_1/2*e^(t/2) + c_2/4*e^(t/4)

Imposing the initial conditions:

y(1) = 4 = c_1e^(1/2)+c_2e^(1/4)

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

c_1 = 4e^(-1/2)-c_2*e^(-1/4)

c_2 = 2e^(-1/4)

c_1 = 2e^(-1/2)

y(t) = 2e^(t/2-1/2)+2e^(t/4-1/4).

Taking the appropriate limits:

lim(y(t),t,-∞) = 0; lim(y(t),t,∞) = ∞

The solution approaches 0 as t approaches -∞, and approaches ∞ as t approaches ∞.

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary): OK

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Self-critique rating: OK

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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” + kr’/m - ω^2r = 0;

I’m going to shorten the process, and jump straight to the characteristic polynomial.

λ^2 + kλ/m - ω^2λ = 0;

Using the quadratic formula:

λ = -k/(2m) ± sqrt((k/m)^2+4ω^2)/2;

r_g = Ae^(-k/(2m)+sqrt((k/m)^2+4ω^2)/2*t) + Be^(-k/(2m)-sqrt((k/m)^2 + 4ω^2)/2*t);

Given additional information:

ω = 20 rev/min = 2pi/3 rad/s

v_0 = 1 cm/s

k/m = 4 s^(-1) >> k = 4m s^(-1)

we have:

r_g ≈ Ae^(0.9t) + e^(-4.9t)

r_g’ = v(t) = 0.9Ae^(0.9t) - 4.9e^(-4.9t)

If r(0) = 0 and r’(0) = v(0) = 1 cm/s, then

A + B = 0;

0.9A-4.9B = 1

A ≈ 0.1724

B ≈ -0.1724

r(t) ≈ 0.1724e^(0.9t) - 0.1724e^(-4.9t)

r(2) = 1.043

Also, if mass is proportional to volume, m = 4/3*π*(d/2)^3, and if drag is proportional to cross sectional area, k = π(d/2)^2. k/m = 3/(2d), which means as the diameter increases, k/m decreases.

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary): OK

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I can't read the details of your solutions (see previous notes regarding the need to use text editors) but your results and the overall approach to each problem are good.

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