Query 01

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course Mth 279

12:25 pm 6/15

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: 

p(t) = -2

P(t) = -2t +C

y = e^(-(-2t+C))

y = Ce^(2t)

I'm assuming you meant y(1) = 3

3 = Ce^(2(1))

3 = Ce^2

C = 3e^(-2)

combining like terms

y = 3 e^(2t-2) 

 

 

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Given Solution: 

 

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Self-critique (if necessary):

 

 I am still unsure of how e^(p(t)) is related.

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Can you show by taking derivatives that

e^(-integral(p(t) dt) )

is a solution to the equation

y ' + p(t) y = 0?(p(t)) is related.  

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

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Your solution: 

 

 y' - 9/t^2 y = 0

p(t) = -9/t^2

P(t) = 9t^(-1) + C

y = Ce^(9/t)

plugging in initial value

2 = Ce^9

C = 2e^(-9)

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

 

 

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Given Solution: 

 

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Self-critique (if necessary):

 

 

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

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Your solution: 

 

 y' + (2t +1)/(t^2 + t) y = 0

p(t) = (2t+1)/(t^2 + t)

I don't know how to do this integral ???

P(t) = lnt + ln(t+1)

y = C e^(lnt + ln(t+1))

y = C( t + t+ 1)

y = C(2t+1)

IVP

1 = C (0+1)

C = 1

y = 2t + 1

 

 

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Given Solution: 

 

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Self-critique (if necessary):

 

 

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

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Your solution: 

 p(t) = sin(3t)

P(t) = -1/3 cos(3t)

 

 y = Ce^(-1/3 cos(3t))

2 = Ce^(-1/3 cos(0)

2 = Ce^(-1/3)

C = 2e^(1/3)

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

 

 

confidence rating #$&*:

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Given Solution: 

 

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Self-critique (if necessary):

 

 

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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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y' = t^2*y

B, because the graph is of a parabola

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y ' - y = 0

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y' = y

graph A because the graph is of y' = y

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y' - y / t = 0

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C because the y=t line is at a constant slope of 1

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y ' - t y = 0

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F because the slopes are 0 when either equal 0

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y ' + t y = 0

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E because the y = 0 and t = 0 lines all have slopes of 0

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A

B

C

D

E

F

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: 

 p(t) = b

P(t) = bt + C

y = C e^(bt)

2 = C e^(b)

8 = Ce^(3b)

I'm not really sure how to solve these two equations.

C = 2/e^(b)

plugging into second equation

8 = (2e^(-b)) e^(b)

8 = (2 e^(-b+b))

8 = 2e ?

There is no solution???

 

 

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Given Solution: 

 

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Self-critique (if necessary):

 

 

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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' = y'

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

w' - w = 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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p(t) = -1

P(t) = -t

w = C e^(-t)

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This should be C e^t.

The function is C e^-(integral(p(t)).

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

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y+2 = Ce^(-t)

y = Ce^(-t) -2

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

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function: y' - y = 2

y = Ce^(-t) - 2

y' = -Ce^(-t)

-Ce^(-t) -[ Ce^(-t) - 2 ] = 2

-Ce^(-t) - Ce^(-t) + 2 = 2

-2 Ce^(-t) = 0

this is not a solution???

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Your solution: 

 I don't understand why we switched from in terms of w to in terms of y, and then back again

and I don't see where I went wrong because my equation is not a solution of y' - y = 2

???

 

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Just a simple error in one step; check my note.

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confidence rating #$&*:

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Given Solution: 

 

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Self-critique (if necessary):

 

 

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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_0 = 1

p(t) = -b

P(t) = -bt

y = Ce^(-bt)

1 = C e^0

C = 1

y = e^(-bt)

plugging in (1, 2)

2 = e^(-b)

ln2 = -b

b = -ln2

confidence rating #$&*:

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Given Solution: 

 

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Self-critique (if necessary):

 

 

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