query 20

#$&*

course Mth 279

4/12 6

Query 20 Differential Equations

********************************************* Question: Using variation of parameters, solve the equation y '' + y = sec(t), -pi/2 < t < pi/2. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Complimentary solution y_c(t) = Acos(t) + Bsin(t), with y_1 = cos(t), y_2 = sin(t) Wronskian of fns is W = cos^2(t) + sin^2(t) = 1 Particular solution is y_p(t) = -cos(t)* Int( (sin(t)*sec(t))/1 dt) + sin(t) Int( (cos(t)*sec(t))/1 dt) =-cos(t)* Int(sin(t)/cos(t) dt) + sin(t) Int( cos(t)/cos(t) dt) =-cos(t)*-ln|cos(t)| + sin(t)*t^2/2 +

@& The integral of cos(t) / cos(t) = 1 is t, not t^2 / 2.*@

General solution is y(t) = Acos(t) + Bsin(t) + cos(t)ln|cos(t)| + sin(t)*t^2/2 ???I know everything is defined but ln(0), so how would that effect our our solution?????

@& The domain of the solution is -pi < t < pi/2.

The cosine is nonzero over this domain. So that isn't a problem.*@

@& Variation of parameters starts from the assumption of a particular solution y_p = u_1 y_1 + u_2 y_2, where y_1 and y_2 are solutions of the homogeneous equation. Making the assumption that u_1 ' y_1 + u_2 ' y_2 = 0 we run through the details and get a second equation u_1 ' y_1 ' + u_2 ' y_2 ' = g, where g is the nonhomogeneous term. Putting our two equations together we get the equation

W(t) * [u_1 ' ; u_2 ' ] = [0; g]

Inverting the Wronskian [y_1, y_2; y_1 ' , y_2 ' ] we get [y_2 ' , - y_2; - y_1 ' , y_1] / || W(t) ||, leading to

[ u_1 '; u_2 ' ] = W^-1(t) * [0; g] = [- y_2 g, y_1 g ] / || W(t) ||

so that

u_1 = integral ( - y_2 g ) u_2 = integral (y_1 g).

For this problem, using the easily-found solutions for the homogeneous equation, namely y_1 = cos(t) and y_2 = sin(t), we get

u_1 = integral ( -sin(t) sec(t) dt) = integral (-sin(t) / cos(t) dt) = - ln | cos(t) | u_2 = integral( cos(t) sec(t) dt) = integral( 1 dt) = t

Thus our particular solution is

y_P = -ln | cos(t) | * cos(t) + t sin(t).

Running through some of the details in this case:

The homogeneous equation yields fundamental set

{cos(t), sin(t)}

so the complementary solution is

y_C (t) = A cos(t) + B sin(t).

We seek a particular solution of the form

y_P = u_1 cos(t) + u_2 sin(t).

y_P ' = u_1 ' cos(t) + u_2 ' sin(t) - u_1 sin(t) + u_2 cos(t).

We will have two simultaneous equations to solve for u_1 and u_2, so we can specify the first to be

u_1 ' cos(t) + u_2 ' sin(t) = 0.

With this assumption, we have

y_P ' = -u_1 sin(t) + u_2 cos(t)

so that

y_P '' = - u_1 ' sin(t) + u_2 ' cos(t) - u_1 cos(t) - u_2 sin(t).

Substituting this into the given equation we get

- u_1 ' sin(t) + u_2 ' cos(t) - u_1 cos(t) - u_2 sin(t) + u_1 cos(t) + u_2 sin(t) = sec(t).

Much of this equation simplifies to zero. Specifically, it should be clear that - u_1 cos(t) - u_2 sin(t) + u_1 cos(t) + u_2 sin(t) = 0. (This is easy to see by just pairing up the terms; the reason it works out this way is that sin(t) and cos(t) are solutions to the homogeneous equation y '' + y ' = 0).

Thus our previous simplfying assumption and the present equation give us

u_1 ' cos(t) + u_2 ' sin(t) = 0

u_1 ' sin(t) + u_2 ' cos(t) = sec(t)

In matrix form this is

[ cos(t), sin(t); sin(t), cos(t) ] * [u_1 ', u_2 ' ] = [0; sec(t)]

Inverting the matrix and multiplying both sides of the equation by the inverse matrix we get expressions for u_1 ' and u_2 ', which we can integrate to obtain u_1 and u_2.

*@

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: Complimentary solution y_c(t) = Acos(3t) + Bsin(3t), with y_1 = cos(3t), y_2 = sin(3t)

@& The homogeneous solutions are cos(6 t) and sin(6 t).*@

Wronskian of fns is W = 3cos^2(3t) + 3sin^2(3t) = 3(1) = 3 Particular solution is y_p(t) = -cos(3t)* Int( (sin(3t)*csc^3(6t))/3 dt) + sin(3t) Int( (cos(3t)*csc^3(6t))/3 dt) =-cos(3t)* Int(sin(3t)/(3*sin^3(6t)) dt) + sin(t) Int( cos(3t)/3sin^3(6t) dt) ???I don’t believe you can calculate the Integral of this, but it’s 3 in the am so I might be overlooking something really obvious??????????

@& The problem is just a careless error in finding the complementary solution. No big deal and not difficult to fix.

You have the procedure down. Just be careful about detail (I should talk about that).*@

General solution is y(t) = Acos(t) + Bsin(t) - cos(3t)* Int(sin(3t)/(3*sin^3(6t)) dt) + sin(t) Int( cos(3t)/3sin^3(6t) dt)

confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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

Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I’m not so sure I did this problem correctly???????? ------------------------------------------------ Self-critique rating: "

@& You have the procedure; just a few errors in detail. Check my notes.*@

query 20

#$&*

course Mth 279

4/12 6

Query 20 Differential Equations

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

Question: Using variation of parameters, solve the equation

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Complimentary solution

y_c(t) = Acos(t) + Bsin(t), with y_1 = cos(t), y_2 = sin(t)

Wronskian of fns is

W = cos^2(t) + sin^2(t) = 1

Particular solution is

y_p(t) = -cos(t)* Int( (sin(t)*sec(t))/1 dt) + sin(t) Int( (cos(t)*sec(t))/1 dt)

=-cos(t)* Int(sin(t)/cos(t) dt) + sin(t) Int( cos(t)/cos(t) dt)

=-cos(t)*-ln|cos(t)| + sin(t)*t^2/2 +

General solution is

y(t) = Acos(t) + Bsin(t) + cos(t)ln|cos(t)| + sin(t)*t^2/2

???I know everything is defined but ln(0), so how would that effect our our solution?????

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:

Complimentary solution

y_c(t) = Acos(3t) + Bsin(3t), with y_1 = cos(3t), y_2 = sin(3t)

Wronskian of fns is

W = 3cos^2(3t) + 3sin^2(3t) = 3(1) = 3

Particular solution is

y_p(t) = -cos(3t)* Int( (sin(3t)*csc^3(6t))/3 dt) + sin(3t) Int( (cos(3t)*csc^3(6t))/3 dt)

=-cos(3t)* Int(sin(3t)/(3*sin^3(6t)) dt) + sin(t) Int( cos(3t)/3sin^3(6t) dt)

???I don’t believe you can calculate the Integral of this, but it’s 3 in the am so I might be overlooking something really obvious??????????

General solution is

y(t) = Acos(t) + Bsin(t) - cos(3t)* Int(sin(3t)/(3*sin^3(6t)) dt) + sin(t) Int( cos(3t)/3sin^3(6t) dt)

confidence rating #$&*:

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

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

Given Solution:

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

Self-critique (if necessary):

I’m not so sure I did this problem correctly????????

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

Self-critique rating:

"

Self-critique (if necessary):

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

Self-critique rating:

________________________________________

#$&*

04-12-2011

query 20

#$&*

course Mth 279

4/12 6

Query 20 Differential Equations

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

Question: Using variation of parameters, solve the equation

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Complimentary solution

y_c(t) = Acos(t) + Bsin(t), with y_1 = cos(t), y_2 = sin(t)

Wronskian of fns is

W = cos^2(t) + sin^2(t) = 1

Particular solution is

y_p(t) = -cos(t)* Int( (sin(t)*sec(t))/1 dt) + sin(t) Int( (cos(t)*sec(t))/1 dt)

=-cos(t)* Int(sin(t)/cos(t) dt) + sin(t) Int( cos(t)/cos(t) dt)

=-cos(t)*-ln|cos(t)| + sin(t)*t^2/2 +

@& The integral of cos(t) / cos(t) = 1 is t, not t^2 / 2.*@

General solution is

y(t) = Acos(t) + Bsin(t) + cos(t)ln|cos(t)| + sin(t)*t^2/2

???I know everything is defined but ln(0), so how would that effect our our solution?????

@& The domain of the solution is -pi < t < pi/2.

The cosine is nonzero over this domain. So that isn't a problem.*@

@&

Variation of parameters starts from the assumption of a particular solution y_p = u_1 y_1 + u_2 y_2, where y_1 and y_2 are solutions of the homogeneous equation. Making the assumption that u_1 ' y_1 + u_2 ' y_2 = 0 we run through the details and get a second equation u_1 ' y_1 ' + u_2 ' y_2 ' = g, where g is the nonhomogeneous term. Putting our two equations together we get the equation

W(t) * [u_1 ' ; u_2 ' ] = [0; g]

Inverting the Wronskian [y_1, y_2; y_1 ' , y_2 ' ] we get [y_2 ' , - y_2; - y_1 ' , y_1] / || W(t) ||, leading to

[ u_1 '; u_2 ' ] = W^-1(t) * [0; g] = [- y_2 g, y_1 g ] / || W(t) ||

so that

u_1 = integral ( - y_2 g )

u_2 = integral (y_1 g).

For this problem, using the easily-found solutions for the homogeneous equation, namely y_1 = cos(t) and y_2 = sin(t), we get

u_1 = integral ( -sin(t) sec(t) dt) = integral (-sin(t) / cos(t) dt) = - ln | cos(t) |

u_2 = integral( cos(t) sec(t) dt) = integral( 1 dt) = t

Thus our particular solution is

y_P = -ln | cos(t) | * cos(t) + t sin(t).

Running through some of the details in this case:

The homogeneous equation yields fundamental set

{cos(t), sin(t)}

so the complementary solution is

y_C (t) = A cos(t) + B sin(t).

We seek a particular solution of the form

y_P = u_1 cos(t) + u_2 sin(t).

y_P ' = u_1 ' cos(t) + u_2 ' sin(t) - u_1 sin(t) + u_2 cos(t).

We will have two simultaneous equations to solve for u_1 and u_2, so we can specify the first to be

u_1 ' cos(t) + u_2 ' sin(t) = 0.

With this assumption, we have

y_P ' = -u_1 sin(t) + u_2 cos(t)

so that

y_P '' = - u_1 ' sin(t) + u_2 ' cos(t) - u_1 cos(t) - u_2 sin(t).

Substituting this into the given equation we get

- u_1 ' sin(t) + u_2 ' cos(t) - u_1 cos(t) - u_2 sin(t) + u_1 cos(t) + u_2 sin(t) = sec(t).

Much of this equation simplifies to zero. Specifically, it should be clear that - u_1 cos(t) - u_2 sin(t) + u_1 cos(t) + u_2 sin(t) = 0. (This is easy to see by just pairing up the terms; the reason it works out this way is that sin(t) and cos(t) are solutions to the homogeneous equation y '' + y ' = 0).

Thus our previous simplfying assumption and the present equation give us

u_1 ' cos(t) + u_2 ' sin(t) = 0

u_1 ' sin(t) + u_2 ' cos(t) = sec(t)

In matrix form this is

[ cos(t), sin(t); sin(t), cos(t) ] * [u_1 ', u_2 ' ] = [0; sec(t)]

Inverting the matrix and multiplying both sides of the equation by the inverse matrix we get expressions for u_1 ' and u_2 ', which we can integrate to obtain u_1 and u_2.

*@

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:

Complimentary solution

y_c(t) = Acos(3t) + Bsin(3t), with y_1 = cos(3t), y_2 = sin(3t)

@& The homogeneous solutions are cos(6 t) and sin(6 t).*@

Wronskian of fns is

W = 3cos^2(3t) + 3sin^2(3t) = 3(1) = 3

Particular solution is

y_p(t) = -cos(3t)* Int( (sin(3t)*csc^3(6t))/3 dt) + sin(3t) Int( (cos(3t)*csc^3(6t))/3 dt)

=-cos(3t)* Int(sin(3t)/(3*sin^3(6t)) dt) + sin(t) Int( cos(3t)/3sin^3(6t) dt)

???I don’t believe you can calculate the Integral of this, but it’s 3 in the am so I might be overlooking something really obvious??????????

@& The problem is just a careless error in finding the complementary solution. No big deal and not difficult to fix.

You have the procedure down. Just be careful about detail (I should talk about that).*@

General solution is

y(t) = Acos(t) + Bsin(t) - cos(3t)* Int(sin(3t)/(3*sin^3(6t)) dt) + sin(t) Int( cos(3t)/3sin^3(6t) dt)

confidence rating #$&*:

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

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

Given Solution:

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

Self-critique (if necessary):

I’m not so sure I did this problem correctly????????

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

Self-critique rating:

"

@& You have the procedure; just a few errors in detail. Check my notes.*@

query 20

#$&*

course Mth 279

4/12 6

Query 20 Differential Equations

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

Question: Using variation of parameters, solve the equation

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Complimentary solution

y_c(t) = Acos(t) + Bsin(t), with y_1 = cos(t), y_2 = sin(t)

Wronskian of fns is

W = cos^2(t) + sin^2(t) = 1

Particular solution is

y_p(t) = -cos(t)* Int( (sin(t)*sec(t))/1 dt) + sin(t) Int( (cos(t)*sec(t))/1 dt)

=-cos(t)* Int(sin(t)/cos(t) dt) + sin(t) Int( cos(t)/cos(t) dt)

=-cos(t)*-ln|cos(t)| + sin(t)*t^2/2 +

General solution is

y(t) = Acos(t) + Bsin(t) + cos(t)ln|cos(t)| + sin(t)*t^2/2

???I know everything is defined but ln(0), so how would that effect our our solution?????

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:

Complimentary solution

y_c(t) = Acos(3t) + Bsin(3t), with y_1 = cos(3t), y_2 = sin(3t)

Wronskian of fns is

W = 3cos^2(3t) + 3sin^2(3t) = 3(1) = 3

Particular solution is

y_p(t) = -cos(3t)* Int( (sin(3t)*csc^3(6t))/3 dt) + sin(3t) Int( (cos(3t)*csc^3(6t))/3 dt)

=-cos(3t)* Int(sin(3t)/(3*sin^3(6t)) dt) + sin(t) Int( cos(3t)/3sin^3(6t) dt)

???I don’t believe you can calculate the Integral of this, but it’s 3 in the am so I might be overlooking something really obvious??????????

General solution is

y(t) = Acos(t) + Bsin(t) - cos(3t)* Int(sin(3t)/(3*sin^3(6t)) dt) + sin(t) Int( cos(3t)/3sin^3(6t) dt)

confidence rating #$&*:

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

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

Given Solution:

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

Self-critique (if necessary):

I’m not so sure I did this problem correctly????????

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

Self-critique rating:

"

Self-critique (if necessary):

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

Self-critique rating:

________________________________________

#$&*

04-12-2011

query 20

#$&*

course Mth 279

4/12 6

Query 20 Differential Equations

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

Question: Using variation of parameters, solve the equation

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Complimentary solution

y_c(t) = Acos(t) + Bsin(t), with y_1 = cos(t), y_2 = sin(t)

Wronskian of fns is

W = cos^2(t) + sin^2(t) = 1

Particular solution is

y_p(t) = -cos(t)* Int( (sin(t)*sec(t))/1 dt) + sin(t) Int( (cos(t)*sec(t))/1 dt)

=-cos(t)* Int(sin(t)/cos(t) dt) + sin(t) Int( cos(t)/cos(t) dt)

=-cos(t)*-ln|cos(t)| + sin(t)*t^2/2 +

@& The integral of cos(t) / cos(t) = 1 is t, not t^2 / 2.*@

General solution is

y(t) = Acos(t) + Bsin(t) + cos(t)ln|cos(t)| + sin(t)*t^2/2

???I know everything is defined but ln(0), so how would that effect our our solution?????

@& The domain of the solution is -pi < t < pi/2.

The cosine is nonzero over this domain. So that isn't a problem.*@

@&

Variation of parameters starts from the assumption of a particular solution y_p = u_1 y_1 + u_2 y_2, where y_1 and y_2 are solutions of the homogeneous equation. Making the assumption that u_1 ' y_1 + u_2 ' y_2 = 0 we run through the details and get a second equation u_1 ' y_1 ' + u_2 ' y_2 ' = g, where g is the nonhomogeneous term. Putting our two equations together we get the equation

W(t) * [u_1 ' ; u_2 ' ] = [0; g]

Inverting the Wronskian [y_1, y_2; y_1 ' , y_2 ' ] we get [y_2 ' , - y_2; - y_1 ' , y_1] / || W(t) ||, leading to

[ u_1 '; u_2 ' ] = W^-1(t) * [0; g] = [- y_2 g, y_1 g ] / || W(t) ||

so that

u_1 = integral ( - y_2 g )

u_2 = integral (y_1 g).

For this problem, using the easily-found solutions for the homogeneous equation, namely y_1 = cos(t) and y_2 = sin(t), we get

u_1 = integral ( -sin(t) sec(t) dt) = integral (-sin(t) / cos(t) dt) = - ln | cos(t) |

u_2 = integral( cos(t) sec(t) dt) = integral( 1 dt) = t

Thus our particular solution is

y_P = -ln | cos(t) | * cos(t) + t sin(t).

Running through some of the details in this case:

The homogeneous equation yields fundamental set

{cos(t), sin(t)}

so the complementary solution is

y_C (t) = A cos(t) + B sin(t).

We seek a particular solution of the form

y_P = u_1 cos(t) + u_2 sin(t).

y_P ' = u_1 ' cos(t) + u_2 ' sin(t) - u_1 sin(t) + u_2 cos(t).

We will have two simultaneous equations to solve for u_1 and u_2, so we can specify the first to be

u_1 ' cos(t) + u_2 ' sin(t) = 0.

With this assumption, we have

y_P ' = -u_1 sin(t) + u_2 cos(t)

so that

y_P '' = - u_1 ' sin(t) + u_2 ' cos(t) - u_1 cos(t) - u_2 sin(t).

Substituting this into the given equation we get

- u_1 ' sin(t) + u_2 ' cos(t) - u_1 cos(t) - u_2 sin(t) + u_1 cos(t) + u_2 sin(t) = sec(t).

Much of this equation simplifies to zero. Specifically, it should be clear that - u_1 cos(t) - u_2 sin(t) + u_1 cos(t) + u_2 sin(t) = 0. (This is easy to see by just pairing up the terms; the reason it works out this way is that sin(t) and cos(t) are solutions to the homogeneous equation y '' + y ' = 0).

Thus our previous simplfying assumption and the present equation give us

u_1 ' cos(t) + u_2 ' sin(t) = 0

u_1 ' sin(t) + u_2 ' cos(t) = sec(t)

In matrix form this is

[ cos(t), sin(t); sin(t), cos(t) ] * [u_1 ', u_2 ' ] = [0; sec(t)]

Inverting the matrix and multiplying both sides of the equation by the inverse matrix we get expressions for u_1 ' and u_2 ', which we can integrate to obtain u_1 and u_2.

*@

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:

Complimentary solution

y_c(t) = Acos(3t) + Bsin(3t), with y_1 = cos(3t), y_2 = sin(3t)

@& The homogeneous solutions are cos(6 t) and sin(6 t).*@

Wronskian of fns is

W = 3cos^2(3t) + 3sin^2(3t) = 3(1) = 3

Particular solution is

y_p(t) = -cos(3t)* Int( (sin(3t)*csc^3(6t))/3 dt) + sin(3t) Int( (cos(3t)*csc^3(6t))/3 dt)

=-cos(3t)* Int(sin(3t)/(3*sin^3(6t)) dt) + sin(t) Int( cos(3t)/3sin^3(6t) dt)

???I don’t believe you can calculate the Integral of this, but it’s 3 in the am so I might be overlooking something really obvious??????????

@& The problem is just a careless error in finding the complementary solution. No big deal and not difficult to fix.

You have the procedure down. Just be careful about detail (I should talk about that).*@

General solution is

y(t) = Acos(t) + Bsin(t) - cos(3t)* Int(sin(3t)/(3*sin^3(6t)) dt) + sin(t) Int( cos(3t)/3sin^3(6t) dt)

confidence rating #$&*:

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

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

Given Solution:

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

Self-critique (if necessary):

I’m not so sure I did this problem correctly????????

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

Self-critique rating:

"

@& You have the procedure; just a few errors in detail. Check my notes.*@