Query 02

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course Mth 279

6/15 3:09 pm

Solve each equation:*********************************************

Question:  1.  y ' + y = 3

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Your solution: 

 p(t) = 1

P(t) = t

u = e^(t)

e^(t) y' + e^(t) y = 3e^(t)

integrate both sides

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

y = 3 + C e^(-t)

 

 

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Given Solution: 

 

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Self-critique (if necessary):

 

 

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Question: 

2.  y ' + t y = 3 t

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Your solution: 

 p(t) = t

P(t) = t^2/2

u = e^(t^2/2)

e^(t^2/2)y' + e^(t^2/2) ty = 3t e^(t^2/2)

integrate both sides. How do you integrate the right side??? I do not understand how to get the answer.

I tried to do integration by parts but it got terribly messy and had 5 terms

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

y = 3 + Ce^(t^2/2)

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The right-hand side is integrated by the u substitution u = t^2 / 2. Integration by part doesn't work for this function, except by a breakdown that's equivalent to the u-substitution.

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confidence rating #$&*:

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Given Solution: 

 

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Self-critique (if necessary):

 Where can I find a refresher of integration by parts???

 

 

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Self-critique rating:

@&

It looks like you probably had a decent refresher in trying to integrate that function using the method.

However I always recommend Khan Academy. Paul's Notes are also very good. Both are very easy to find.

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Question: 

3.  y ' - 4 y = sin(2 t)

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Your solution: 

 

 p(t) = -4

P(t) = -4t

u = e^(-4t)

e^(-4t) y' - 4e^(-4t) y = e^(-4t) * sin(2t)

integrate both sides. I also don't know how to do this integral. I know it can't be done by tabular or integration by parts. ???

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

y = -1/10 (2sin(2t) + cos(2t)) + Ce^(4t)

 

confidence rating #$&*:

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Given Solution: 

 

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Self-critique (if necessary):

 I just don't know how to integrate

e^(-4t)sin(2t) ???

 

@&

Two integrations by parts, each with u = e^(-4 t) and du = -4 e^(-4 t), will lead you to an expression which contains a multiple of the original integral. You can set this expression equal to the original integral and solve for the original integral.

This is called a reduction formula. Part of a standard first-year introduction to integration.

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Question: 

4.  y ' + y = e^t, y (0)  = 2

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Your solution: 

 p(t) = 1

P(t) = t

u = e^(t)

 

 e^(t) y' + e^(t) y = e^(2t)

integrate both sides

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

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

plugging in IVP

2 = 1/2 e^(0) + C e^(0)

C = 1.5

y = 1/2 e^(t) + 1.5e^(-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: 

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

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Your solution: 

 

 p(t) = 3

P(t) = 3t

u = e^(3t)

e^(3t) y' + 3e^(3t) y = 3e^(3t) + 2te^(3t) + e^(t)e^(3t)

integrate both sides, using tabular on the second term

e^(3t) y = e^(3t) + [ 6te^(3t) -18e^(3t) ] + 1/4 e^(4t) +C

y = 1 + 6t - 18 + 1/4 e^(t) + Ce^(-3t)

y = 6t - 17 + .25e^(t) + Ce^(-3t)

plugging in IVP

e^2 = 6 - 17 + .25e + C e^(-3)

11 -.25 e + e^2 = Ce^(-3)

11e^(3) - .25e^(4) + e^(5) = C

 Can you simplify this???

@&

You could factor out e^3, though that's not much of a simplification in this case.

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y = 6t - 17 + .25e^(t) + e^(-3t) *[ 11e^(3) - .25e^(4) + e^(5) ]

 

 

confidence rating #$&*:

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Given Solution: 

 

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Self-critique (if necessary):

 

 

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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: 

 going backwards

u = e^(t^2)

P(t) = t^2

p(t) = 2t

my guess is g(t) = 1, but I'm not too sure why

do everything I normally do, but backwards

multiply right side by u

C + e^(t^2)

take the derivative

2t e^(t^2)

divide by u

2t

just kidding, g(t) = 2t and that is my final answer

 

 

confidence rating #$&*:

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Given Solution: 

 

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Self-critique (if necessary):

 

 

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#*&!

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You had a couple of problems with integration but your work was otherwise very good.

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Check my notes.

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