Query_01

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course Mth 279

2/14

Section 2.2*********************************************

Question: 1. Solve the following equations with the given initial conditions:

1. y ' - 2 y = 0, y(1) - 3

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Your solution:

y ' - 2 y = 0

dy/dt - 2y = 0

dy/dt = 2y

dy = 2y dt

dy/y = 2 dt

int(dy/y) = 2 * int(dt)

ln(y) = 2t + c

y = e^(2t + c)

y = c * e^(2t)

y(1) = 3

y(1) = c * e^(2(1)) = 3

c = 3/e^2

y = (3/e^2) * e^(2t)

y = 3e^(2t) /e^2

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: 2. t^2 y ' - 9 y = 0, y(1) = 2.

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Your solution:

t^2 y ' - 9 y = 0

t^2 dy/dt = 9 y

dy/y = 9 dt/t^2

int (dy/y) = 9 * int (dt/t^2)

ln(y) = -9/t + c

y(t) = e^(c - 9/t)

y(t) = c/e^(9/t)

y(1) = 2

y(1) = c/e^(9/1) = 2

c = 2e^9

y(t) = 2e^9 / e^(9/t)

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: 3. (t^2 + t) y' + (2t + 1) y = 0, y(0) = 1.

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Your solution:

(t^2 + t) y' + (2t + 1) y = 0

dy/dt = -(2t + 1) y /(t^2 + t)

dy/y = -(2t + 1)/(t^2 + t) dt

int (dy/y) = -int((2t + 1)/(t^2 + t) dt)

ln(y) = -ln(t^2 + t) + c

y(t) = c/(t^2 + t)

y(0) = 1

y(0) = c/(0^2 + 0) = c/0 = 1

cannot divide by zero, so c is undefined by this condition.

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: 4. y ' + sin(3 t) y = 0, y(0) = 2.

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Your solution:

y ' + sin(3 t) y = 0

dy/dt + sin(3 t) y = 0

dy/dt = -sin(3 t) dt

int (dy/dt) = - int (sin(3 t) dt)

ln(y) = cos(3t) /3 + c

y(t) = e^(cos(3t) /3 + c)

y(t) = c * e^(cos(3t) /3)

y(0) = 2

y(0) = c * e^(cos(3* 0) /3) = 2

y(0) = c * e^(1/3) = 2

c = 2/e^(1/3)

y(t) = 2/e^(1/3) * e^(cos(3t) /3)

y(t) = 2e^(cos(3t) /3) /e^(1/3)

confidence rating #$&*:

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3

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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: 5. Match each equation with one of the direction fields shown below, and explain why you chose as you did.

y ' - t^2 y = 0

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Using the method y = e^(-int(p(t)dt)) when in the form y' + p(t)y = 0

y = c*e^(t^3/3)

E

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y ' - y = 0

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Using the method y = e^(-int(p(t)dt)) when in the form y' + p(t)y = 0

y = c*e^t

A

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y' - y / t = 0

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Using the method y = e^(-int(p(t)dt)) when in the form y' + p(t)y = 0

y = t + c

C

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y ' - t y = 0

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Using the method y = e^(-int(p(t)dt)) when in the form y' + p(t)y = 0

y = c*e^(t^2/2)

B

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y ' + t y = 0

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Using the method y = e^(-int(p(t)dt)) when in the form y' + p(t)y = 0

y = c*e^(-t^2/2)

F

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A

B

C

D

E

F

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Good, but the point of this problem is not to solve the equation but to associate the geometry of the direction field with the equation.

In each case you can rearrange the equation to give you a formula for y '. Evaluating y ' at a grid of y vs. t points and sketching slope segments, you construct the direction field.

You can do this whether the equation is solvable or not, and the process gives you a good picture of the trends of the solution curves.

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6. The graph of y ' + b y = 0 passes through the points (1, 2) and (3, 8). What is the value of b?

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Your solution:

y ' + b y = 0

y(t) = c * e^(-bt)

(t,y)

y(1) =c * e^(-b) = 2

c = 2e^b

y(3) =c * e^(-b(3)) = 8

c = 8e^(3b)

as for figuring out what b is, I am stumped after this.

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The two points give you

2 = c e^(-b * 1)

and

8 = c e^(-b * 8).

If you divide one of the equations by the other you will eliminate c and obtain an equation you can solve for b.

Substituting that value of b into either equation gives you an equation you can solve for c.

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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:

7. The equation y ' - y = 2 is first-order linear, but is not homogeneous.

If we let w(t) = y(t) + 2, then:

What is w ' ?

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w(t) = y(t) + 2

w'(t) = (y(t) + 2)' = y'(t) + (2)' = y'(t)

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What is y(t) in terms of w(t)?

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y(t) = w(t) - 2

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What therefore is the equation y ' - y = 2, written in terms of the function w and its derivative w ' ?

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y ' - y = 2

w'(t) - (w(t) - 2) = 2

w'(t) - w(t) + 2 = 2

w'(t) - w(t) = 0

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Now solve the equation and check your solution:

Solve this new equation in terms of w.

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w(t) = c*e^t

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Substitute y + 2 for w and get the solution in terms of y.

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w(t) = c*e^t

y + 2 = c*e^t

y = c*e^t -2

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Check to be sure this function is indeed a solution to the equation.

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y ' - y = 2

c*e^t - (c*e^t -2) = 2

c*e^t - c*e^t + 2 = 2

2 = 2

solution works!

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Your solution:

confidence rating #$&*:

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3

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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: 8. The graph below is a solution of the equation y ' - b y = 0 with initial condition y(0) = y_0. What are the values of y_0 and b?

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Your solution:

y(t) = c * e^(-bt)

y_0 = 1 -> y intercept.

y(0) = c * e^(-b*0) = 1

c = 1

y(1) = c * e^(-b*1) = 2

y(1) = e^(-b) = 2

b = -ln(2)

confidence rating #$&*:

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

3

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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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Good work. Check my notes.

You can also check the document at

http://vhmthphy.vhcc.edu/tests/shell_01/differential_equations/query_01_solnerngkf0jknj22lkkkls.htm

for further discussion of these solutions. You will probably want to do that for the problem about direction fields.

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