#$&*
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 #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution:
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
I’m not so sure I did this problem correctly????????
------------------------------------------------
Self-critique rating:
"
@& You have the procedure; just a few errors in detail. Check my notes.*@
#$&*
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:
________________________________________
#$&*
04-12-2011
#$&*
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.*@
#$&*
course Mth 279
4/12 6
Query 20 Differential Equations
*********************************************
Question: Using variation of parameters, solve the equation
y '' + y = sec(t), -pi/2 < t < pi/2.
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
Complimentary solution
y_c(t) = Acos(t) + Bsin(t), with y_1 = cos(t), y_2 = sin(t)
Wronskian of fns is
W = cos^2(t) + sin^2(t) = 1
Particular solution is
y_p(t) = -cos(t)* Int( (sin(t)*sec(t))/1 dt) + sin(t) Int( (cos(t)*sec(t))/1 dt)
=-cos(t)* Int(sin(t)/cos(t) dt) + sin(t) Int( cos(t)/cos(t) dt)
=-cos(t)*-ln|cos(t)| + sin(t)*t^2/2 +
General solution is
y(t) = Acos(t) + Bsin(t) + cos(t)ln|cos(t)| + sin(t)*t^2/2
???I know everything is defined but ln(0), so how would that effect our our solution?????
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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
#$&*
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.*@