Query 12

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course MTH 279

Query 12 Differential Equations*********************************************

Question: A cylinder floating vertically in the water has radius 50 cm, height 100 cm and uniform density 700 kg / m^3. It is raised 10 cm from its equilibrium position and released. Set up and solve a differential which describes its position as a function of clock time.

The same cylinder, originally stationary in its equilibrium position, is struck from above, hard enough to cause its top to come just to but not below the level of the water. Solve the differential equation for its motion with this condition, and use a particular solution to determine its velocity just after being struck.

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Your solution:

y’’ + ω^2*y = 0 ; y(0) = -10 ; y’(0) = 0

r = 50cm, h = 100cm, ρ = 700 kg/m^3, ρ_l = 1000 kg/m^3

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

Substituting:

y’’ = λ^2e^(λt)

λ^2e^(λt) + ω^2e^(λt) = 0

e^(λt) (λ^2 + ω^2) = 0

Since e^(λt) cannot take a value of 0, the 0 must come from the expression in parantheses.

λ^2 = -ω^2

λ = ±iω

λ = -iω >> y_1(t) = c_1*e^(-iωt)

λ = iω >> y_2(t) = c_2*e^(iωt)

y(t) = y_1 + y_2 = c_1*e^(-iωt) + c_2*e^(iωt)

Euler’s Identity: e^(α+iβ) = e^α*cos(β) + i*e^α*sin(β)

y(t) = (c_1+c_2)cos(ωt) + i(-c_1+c_2)sin(ωt)

c_1 = c_1 + c_2; c_2 = i(-c_1+c_2)

y(t) = c_1*cos(ωt) + c_2*sin(ωt)

y(0) = -.10 = c_1*cos(0) + c_2*sin(0)

c_1 = -.10

y’(t) = -c_1*ω*sin(ωt) + c_2*ω*cos(ωt)

y’(0) = 0 = -c_1*ω*sin(0) + c_2*ω*cos(0)

c_2 = 0

y = -.10cos(ωt)

ω^2 = (ρ_l*g)/(ρ*h) = (1000*9.81)/(700*0.1) = 140.14 s^(-2)

y(t) = -.10cos(sqrt(140.14)t)

For part B, using the ω found earlier, we know that one period will be 2π/sqrt(140.14) seconds (or about 0.53s). We can divide this by 4 to find the time to go from a perturbed state back to its equilibrium position (about 0.13s), or vice versa. Using this information, we can determine that y(0) = 0, y(0.13) = 0.3.

y([2π/sqrt(140.14)]/4) = 0.3

y(0) = 0

Using the form from above:

y(t) = c_1cos(ωt) + c_2sin(ωt)

y([2π/sqrt(140.14)]/4) = 0.3 = c_2sin(π/2)

c_2 = 0.3

y(0) = c_1cos(0) + c_2sin(0)

c_1 = 0

y(t) = 0.3sin(ωt) m

y’(t) = 0.3ωcos(ωt)

y’(0) = 0.3*ω ≈ 3.55 m/s

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:

For each of the equations and initial conditions below, find the largest t interval in which a solution is known to exist:

• y '' + y ' + 3 t y = tan(t), y(pi) = 1, y ' (pi) = -1

• t y '' + sin(2 t) / (t^2 - 9) y ' + 2 y = 0, y(1) = 0, y ' (1) = 1.

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Your solution:

a) p(t) = 1, continuous (-∞,∞); q(t) = 3t, continuous (-∞,∞); g(t) = tan(t), continuous except at all t = nπ/2 where n = all odd integers, positive and negative. The largest t interval on which a solution is known to exist is π/2 < t < 3π/2.

b) p(t) = sin(2t)/(t(t^2-9)), continuous except at t = 0,±3; q(t) = 2/t, continuous everywhere except t = 0; g(t) = 0, continuous for all reals. Since the initial conditions are imposed at t = 1, the largest interval on which a solution is known to exist is 0 < t < 3.

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: Decide whether the solution of each of the following equations is increasing or decreasing, and whether each is concave up or concave down in the vicinity of the given initial point:

• y '' + y = - 2 t, y(0) = 1, y ' (0) = -1

• y '' - y = t^2, y(0) = 1, y ' (0) = 1

• y '' - y = - 2 cos(t), y (0) = 1, y ' (0) = 1.

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Your solution:

a) y’’(t) = f(t,y,y’) = -y - 2t; y’’(t_0) = -1, y’_0 = -1 >> solution is concave down and is decreasing

b) I’m going to assume that this was a repeat typo, and say that it is concave down and decreasing as well.

c) y’’(t) = f(t,y,y’) = t^2 + y; y’’(t_0) = 1; y’(t_0) = 1 >> Solution is concave up and is increasing

d) y’’(t) = f(t,y,y’) = -2cos(t) + y; y’’(t_0) = -1; y’(t_0) = 1 >> Solution is concave down and is increasing.

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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