Query 01

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course Mth 279

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: 

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

 p(t)=2

P(t) = 2t

ln(y)=2t+C

C = 3/(e^2)

y(t) = 3*e^(2(t-1))

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confidence rating #$&*:232;**** 

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ok

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Given Solution: 

****

 ok

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Self-critique (if necessary):

****

 ok

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Self-critique rating:

****

3

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

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Your solution: 

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 t^2 y ' - 9 y = 0

y' = 9/(t^2) * y

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

2 = C * e^(-9)

C = 2/e^9

y(t) = 2 * e^(9(1-t^-1))

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confidence rating #$&*:232;****

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

 OK

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Given Solution: 

 

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Self-critique (if necessary):

****

 OK

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Self-critique rating: 3

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

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Your solution: 

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 t^2 + t) y' + (2t + 1) y = 0

 

@&

You need to rearrange this into the form

y ' + p(t) y = 0

and use the method of this section, which dictates that the solution is

y = C e^(integral(p(t) dt) )

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#$&* 

 

 

confidence rating #$&*:232; 

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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: 

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 y ' + sin(3 t) y = 0

y' = -sin(3t)y

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

2 = C * e^(1/3 * cos(0))

C = 2/ e^(1/3)

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

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confidence rating #$&*:232; ok

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

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Given Solution: 

 

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Self-critique (if necessary):

 ok

 

 

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Self-critique rating:

3

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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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Matches E

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y ' - y = 0

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Matches A

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y' - y / t = 0

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Matches C

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y ' - t y = 0

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Unsure

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y ' + t y = 0

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Unsure

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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: 

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 I could not find a problem similar to this one in the book.

I realize that I have been given two points and that I should plug those points into the given equation. I should then use a system of equations to solve for b. I am not exactly sure how to plug in the two points.

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confidence rating #$&*:232; Not Confident

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Given Solution: 

 

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Self-critique (if necessary):

 1

 

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Self-critique rating:

1

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Question: 

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

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I do not understand what you are trying to ask with these questions. Why did we jump from having an equation in the form y ' - y = 2, to having w(t) = y(t) + 2?

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If you answer the questions, you'll be in a position to understand why we do this.

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

What is w ' ?

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You answer this question by taking the derivative of both sides.

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

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You answer this question by solving the equation

w(t) = y(t) + 2

for y(t).

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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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Now solve the equation and check your solution:

Solve this new equation in terms of w.

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

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

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Your solution: 

 

 

confidence rating #$&*:232; 

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

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Given Solution: 

 

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Self-critique (if necessary):

 

 

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Self-critique rating:

@&

The answer to each question is fairly straightforward, but to understand the point unless you need to answer the questions, or attempt to do so.

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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: 

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I do not understand the question at all

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What are the solutions to the equation y ' - b y = 0 and how is the value of b related to the solution curves?

How is the value of y(0) related to the solution curves?

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confidence rating #$&*:232; 

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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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Self-critique (if necessary):

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Self-critique rating:

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Self-critique (if necessary):

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Self-critique rating:

#*&!

@&

You don't seem to be using the methods of the section to answer the questions. It isn't clear how you obtained each solution, but it appears you are solving many of these equations as separable equations. They are separable, but first-order linear are not generally separable, which is why it is necessary to use the specified methods.

See also my notes on a couple of the questions you did not answer.

&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).

Be sure to include the entire document, including my notes.

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