Assignment 14

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course Mth 279

7/15

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:

A)p_w = 1000kg/m^3; p_cyl = 700 kg/m^3; h= 100cm = 1m

Vol difference = -Area * position = -(pir^2)y

mass displaced = -pr^2y

F_net = -p_w(pir^2)y*g

m_cyl = p*pir^2h

F_net = my’’

so the plugging in everything we get

p_cyl*pir^2h*y’’ = -p_w*pir^2gy

y’’ = (-p_w*g)y/(h*p_cyl) = (-1000*9.8)y/(700*10) = -14y

No idea where to go from here or what to do???

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The solution to

y '' = -c y

is

y = A cos(sqrt(c) t) + B sin(sqrt(c) t) )

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You should check out that it is so, then review the materials to see how this solution comes about.

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confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

I took problem to as far as I could work it then I didn’t know what to do to finish/answer the problem.

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Self-critique rating:

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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) y’’ + y’ + 3ty = tan(t)

so p(t) = 1 ; q(t) = 3t and g(t) = tan(t)

p and q are continuous everywhere

g discontinuous at every odd multiple of pi/2

t_0 = pi

so now we need to find the interval where t_0 falls in-between.

we would have intervals of (-3pi/2, -pi/2) , (-pi/2, pi/2), (pi/2, 3pi/2) and so on.

But the only integral that t_0 falls between is

(pi/2, 3pi/2)

B) ty’’ + sin(2t)/(t^2 - 9) y’ + 2y = 0

divide by t

y’’ + sin(2t)/(t(t^2 -9)) y’ + 2y/t = 0

p = sin(2t)/(t(t^2 - 9)); q = 2y/t and g = 0

to find where p and q are discontinuous you find where the denominators are = 0

for p:

t(t^2 - 9) = 0

t = 0; t = 3, and t = -3

for q:

t= 0

so find the largest interval when t = 1

we would have the intervals of:

(-infinity, -3) (-3,0) (0,3) (3,infinity)

so t=1 falls between (0,3)

When if ever would an interval be <= or >= ???

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Good question.

The given initial condition occurs at only one point, so even if (as in the present case) the equation is defined for a split interval, the solution can be defined only in that part of the interval that contains the given point.

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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: 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 = - 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’’ + y = -2t

initial point (0,1) with a slope of -1, so function is decreasing.

y’’ = -1 - 2(0) = -1 with a neg second derivative the function is concave down

B) y’’ -y = t^2

slope = 1 so function is increasing

y’’ = y + t^2 = 1 + 0^2 = 1 so function is concave up

C) y’’ - y = -2cos(t)

slope =1 so increasing

y’’ = 1 - 2cos(0) = 1 - 2(1) = -1 so concave down

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Good.

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