query 20

#$&*

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

@& 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.

*@

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

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).*@

@& 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).*@

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

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

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query 20

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

@& 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.

*@

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

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

@& 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).*@

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

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

query 20

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

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

query 20

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

@& 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.

*@

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

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

@& 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).*@

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

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