Query 20

#$&*

course Mth 279

RESUBMISSON. I submitted this I think last week along with Query 21, I do have the form confirmation, but I think it just got missed somehow. Also, thank you very much for the continued support. I should have a stream of assignments coming your way soon.

8/5 5:34 pm" "Query 20 Differential Equations

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

Question:  Using variation of parameters, solve the equation

y '' + y = sec(t), -pi/2 < t < pi/2.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution: 

y^2 + 1 = 0

y = +-i

yc = c1 cos(t) + c2 sin(t)

yp = y1 u1 + y2 u2

y1u1' + y2 u2' = 0

y1'u1' + y2'u2' = sec(t)

[cos(t) sin(t),

-sin(t) cos(t)] * [u1', u2'] =

[0, sec(t)]

W = cos^2 + sin^2 = 1

u1' = -sin(t) sec(t) = -sin/cos = tan(t)

u1 = ln(sec(t))

u2' = cos(t) sec(t) = 1

u2 = t

yp = cos(t) ln(sec) + t sin(t)

y = c1cos(t) + c2sin(t) + cos ln(sec(t)) + tsin(t)

 

confidence rating #$&*:

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

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

Given Solution: 

 

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

Self-critique (if necessary):

 

 

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

Self-critique rating:

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

Question:  Using variation of parameters, solve the equation

y '' + 36 y = csc^3 ( 6 t ).

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution: 

 y^2 + 36 = 0

y1 = 6i

yc = c1 cos(6t) + c2 sin(6t)

yp = y1u1 + y2u2

y1u1' + y2u2' = 0

y1'u1' + y2'u2' = csc^3(6t)

[cos(6t) sin6(t),

-6sin(6t) 6cos(6t)] * [u1' u2'] =

[0, csc^3(6t)]

6cos^2(6t) + 6 sin^2(6t) = 6

[u1', u2'] = 1/6 * [6cos(6t) -sin(6t),

6sin(6t) cos(6t)]

u1' = -1/6 sin(6t) csc^3(6t)

u2' = 1/6 cos(6t) csc^3(6t)

I don't think I did this integral right

integral [-1/6 csc^2(6t)]

u substitution

u = csc(6t)

du = 6cot(6t)

-1/6 (6cot(6t)) integral[u^2]

-cot(6t) [1/3 u^3]

-1/3 cot(6t)csc^3(6t) = u1

u2 = integral[1/6 cos(6t) csc^3(6t)]

using csc^2 = 1 + cot^2

u2 = integral [ 1/6 cot(6t)(1 + cot^2(6t))

u2 = integral [ 1/6 cot(6t) + cot^4(6t)]

u2 = 1/6 (1/6) csc(6t) + 1/5 cot^5(6t)

y = c1 cos(6t) + c2 sin(6t) + -1/3 cot(6t)csc^3(6t) + 1/36 csc(6t) + 1/5 cot^5(6t)

 

 

confidence rating #$&*:

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

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

Given Solution: 

 

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

Self-critique (if necessary):

 I don't remember how to integrate things like cot^4 or csc^2

 

 

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

Self-critique rating:

@&

You're unlikely to have to integrate anything that tricky.

You would start, though, by playing around with integration by parts.

For example to integrate cos^2(theta) you would let u = cos(theta) and dv = cos(theta) dTheta. The integral that results is a multiple of the integral of sin^2(theta). Your original integral is of cos^2(theta), which is 1 - sin^2(theta). So your left- and right-hand sides contain terms equal to the integral of sin^2(theta).

etc..

*@

"

Self-critique (if necessary):

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

Self-critique rating:

"

Self-critique (if necessary):

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

Self-critique rating:

#*&!

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