Query 2

#$&*

course MTH 279

1/28 4 pm

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

Question: 1. y ' + y = 3

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

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

y' + y = 3

I(t) = e^t

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

(e^t * y)' = 3e^t integrate both sides

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

y = 3 + C/(e^t)

confidence rating #$&*:

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

.............................................

Given Solution:

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

Self-critique (if necessary): OK

------------------------------------------------

Self-critique rating: OK

*********************************************

Question:

2. y ' + t y = 3 t

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

y = 3 + Ce^(-.5 * t^2)

I(t) = e^(.5 * t^2)

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

(e^(.5 * t^2) * y)' = 3 * e^(.5 * t^2) * t integrate by using u substitution

e^(.5 * t^2) * y = 3 * e^(.5 * t^2) + C

y = 3 + C/[e^(.5 * t^2)]

confidence rating #$&*:

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

.............................................

Given Solution:

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

Self-critique (if necessary): OK

------------------------------------------------

Self-critique rating: OK

*********************************************

Question:

3. y ' - 4 y = sin(2 t)

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

y = .25cos(2t) - .5sin(2t)

I(t) = e^(-4t)

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

Then I did integration by parts twice on the integral, first time let u =e^(-4t) and dv= sin(2t) dt. Second time, let u = e^(-4t) and dv= cos(2t) dt.

@&

The task, as you know, is to integrate the right-hand side. The twofold integration by parts is indeed the way to handle that integral.

*@

confidence rating #$&*:

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

.............................................

Given Solution:

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

Self-critique (if necessary): I wasn't sure how to handle that integral, so I used integration by parts to try to come up with an equivalent expression and I don't know if I handled it properly. I was tempted to leave to answer as y = integral [ e^(-4t) * sin(2t) dt] /e^(-4t), but that seemed...wrong.

------------------------------------------------

Self-critique rating: 3

*********************************************

Question:

4. y ' + y = e^t, y (0) = 2

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

y = .5e^t + 1.5e^-t

I(t) = e^t

(e^t * y)' = e^(2t)

e^t * y = .5e^(2t) + C integrate both sides. I used u sub and let u = 2t

y = .5e^t + Ce^-t

y(0) = 2 = .5 + C

1.5 = C

confidence rating #$&*:

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

.............................................

Given Solution:

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

Self-critique (if necessary): OK

------------------------------------------------

Self-critique rating: OK

*********************************************

Question:

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

y = (7/9) + (2/3)t + (1/4)e^t + e^3 * [e^2 - (52 + 9e)/36 ]/(e^(3t) )

I(t) = e^(3t)

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

Then integrated (I used int. by parts on the 2te^(3t) term.

y = (7/9) + (2/3)t + (1/4)e^t +C/(e^(3t) )

y(1) = e^2 = (7/9) + (2/3)*1 + (1/4)e^1 +C/(e^(3*1)

C = e^3 * [e^2 - (52 + 9e)/36 ]

confidence rating #$&*:

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

.............................................

Given Solution:

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

Self-critique (if necessary): OK

------------------------------------------------

Self-critique rating:

*********************************************

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)?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

p(t) = 2t

g(t) = 2t

I basically worked to problem backwards.

y = C e^(-t^2) + 1 multiply thru by e^(t^2)

e^(t^2)y = C + e^(t^2) take the derivative of each side.

(e^(t^2)y)' = 2te^(t^2) expand the left hand side.

e^(t^2) 8 y' + 2t * e^(t^2) * y = 2te^(t^2) divide through by the integrating factor.

y' *2ty = 2t

confidence rating #$&*:

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

.............................................

Given Solution:

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

Self-critique (if necessary): OK

------------------------------------------------

Self-critique rating: OK"

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