Query 02

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course Mth 279

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Question:  1.  y ' + y = 3

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Your solution: 

 ****

 y ' + y = 3

y' = 3-y

integral ((y'/(3-y)) dt = integral (1) dt

-ln(3-y) = t + C

y(t) = -e^-(x+c) + 3

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This is not the method to use for this section.

The equation is solved using integrating factor e^(integral(p(t) dt) ) = e^t.

Multiplying through by this integrating factor we get the equation

e^t y ' + e^t y = 3 e^t.

The left-hand side is the differential of e^t * y, so when we integrate both sides we get

e^t * y = integral(3 e^t)

e^t * y = 3 e^t + c

y = (3 e^t + c) / e^t = 3 + c e^(-t).

The solution is therefore

y = 3 + c e^(-t).

This is completely equivalent to your solution. Your solution method, however, is limited to first-order linear equations which happen to be separable. Most are not.

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For example if the present equation was changed very slightly to

y ' + t y = t^2

it would not be separable, but it would still be easily solved by the use of an integrating factor.

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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:

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Question: 

2.  y ' + t y = 3 t

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Your solution: 

****

 y ' + t y = 3 t

 y' / (3-y) = t

take integral of both sides

ln(3-y) = t^2 / 2 + C

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

@&

You are not using the appropriate method to solve this equation. The goal here is to learn to solve first-order linear differential equations, and this goal is not achieved by reducing such an equation to a linear equation.

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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:

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Question: 

3.  y ' - 4 y = sin(2 t)

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Your solution: 

****

 y ' - 4 y = sin(2 t)

y' = sin(2t) + 4y

let u(t) = e^(-4t)

y * e^(-4t) = sin (2t)e^(-4t)

y(t) = -(1/10) * cos(2t) - (1/5) * sin(2t) + C*e^(4t)

@&

e^(-4 t) is certainly the appropriate integrating factor, and this can lead you to the equation

y * e^(-4t) = sin (2t)e^(-4t)

but you haven't indicated the steps that lead from the original equation to this equation.

Nor have you indicated how this new form of the equation leads you to your solution

y(t) = -(1/10) * cos(2t) - (1/5) * sin(2t) + C*e^(4t).

These steps would be expected on a test.

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confidence rating #$&*:232; ok

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Given Solution: 

 

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

 ok

 

 

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Self-critique rating:

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Question: 

4.  y ' + y = e^t, y (0)  = 2

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Your solution: 

 y ' + y = e^t

let u(t) = e^t

y*e^t = (1/2) * e^(2t) + C

C = 3/2

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

confidence rating #$&*:232; ok

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Given Solution: 

 

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

 ok

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Self-critique rating:

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Question: 

5.  y ' + 3 y = 3 + 2 t + e^t, y(1) = e^2

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Your solution: 

 y ' + 3 y = 3 + 2 t + e^t

let u(t) = e^(3t)

y*e^(3t) = e^(3t) + e^(4t)/4 + (2/3) * e^(3t) * t - (2/9) * e^(3t) + C

y(t) = (1/36) * (9*e^t + 24*t + (-52 + 36 * e^2 - 9*e)*e^(-3t+3) + 28)\

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:

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Question: 

6.  The general solution to the equation y ' + p(t) y = g(t) is y = C e^(-t^2) + 1, t > 0.  What are the functions p(t) and g(t)?

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Your solution: 

 Not sure what to do.

I think that I need to somehow know that this equation will have a certain form of solution… and use that to solve for p(t) and g(t).

 

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What is the integrating factor for the equation

y ' + p(t) y = g(t)?

What are the steps in solving the general equation, and what is the form of the general solution?

How does this match up with the form of the specific solution

y = C e^(-t^2) + 1?

*@

 

confidence rating #$&*:232; 1

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Given Solution: 

 

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Self-critique (if necessary):

 1

 

 

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Self-critique rating:"

@&

You have shown an outline of your solutions to a few of these equations which is based on the appropriate method, though in your first couple of problems you did not use the appropriate method. That's not bad.

But you've left out the steps that would really demonstrate that you understand the method.

Check my notes. I'll leave it to you whether you want to submit a revision of some or all of these problems. If you do, use the protocol in the subsequent paragraph; however note that the revision is in this case optional.

&#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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