Query_06

#$&*

course MTH-279

06/21 around 5:56AM.

Query 06 Differential Equations*********************************************

Question: 3.3.4. If the equation is exact, solve the equation (6 t + y^3 ) y ' + 3 t^2 y = 0, with y(2) = 1.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

(6 t + y^3 ) y ' + 3 t^2 y = 0

Since in the form Mdt + Ndy = 0

M = 3 t^2 y

N = (6 t + y^3 )

dM/dy = 3t^2

dN/dt = 6

Since these two are not equal the equation is not exact.

confidence rating #$&*:

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

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

Given Solution:

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

Self-critique (if necessary):

Self-critique rating:

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

Question: If the equation is exact, solve the equation (6 t + 3 t^3 ) y ' + 6 y + 9/2 t^2 y^2 = -t, with y(2) = 1. *

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

(6 t + 3 t^3 ) y ' + 6 y + 9/2 t^2 y^2 + t = 0

Since in the form Mdt + Ndy = 0

M = 6 y + 9/2 t^2 y^2 + t

N = (6 t + 3 t^3 )

dM/dy = 9yt^2 + 6

dN/dt = 9t^2 + 6

These are not equivalent either, so the equation is not exact.

confidence rating #$&*:

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

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

Given Solution:

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

Self-critique (if necessary):

Self-critique rating:

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

Question: 3.3.6. If the equation is exact, solve the equation y ' = - ( y cos(t y) + 1) / (t cos(t y) + 2 y e^y^2) with y(0) = pi.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

y ' = ( y cos(t y) + 1) / (t cos(t y) + 2 y e^y^2)

(t cos(t y) + 2 y e^y^2)y ' = ( y cos(t y) + 1)

(t cos(t y) + 2 y e^y^2)y ' - ( y cos(t y) + 1) = 0

M = - ( y cos(t y) + 1) = -y cos(t y) - 1

N = (t cos(t y) + 2 y e^y^2)

dM/dy = tysin(ty)-cos(ty)

dN/dt = cos(ty) - tysin(ty)

These are close but not equivalent either.

confidence rating #$&*:

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

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

Given Solution:

Self-critique (if necessary):

Self-critique rating:

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

Question: 3.3.10. If N(y, t) * y ' + t^2 + y^2 sin(t) = 0 is exact, what is the most general possible form of N(y, t)?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

N(y, t) * y ' + t^2 + y^2 sin(t) = 0

M = t^2 + y^2 sin(t)

N = ?

dM/dy = 2y sin(t)

Int(2y sin(t)) dt = -2ycos(t)

N = -2ycos(t)

Confidence rating: 2

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

Given Solution:

Self-critique (if necessary):

Self-critique rating:

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

Question: 3.3.12. If y = -t - sqrt( 4 - t^2 ) is the solution of the differential equation (y + a t) y ' + (a y + b t) = 0 with initial condition y(0) = y_0, what are the values of a, b and y_0?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

y = -t - sqrt( 4 - t^2 )

Solution: (y + a t) y ' + (a y + b t) = 0

First we can solve the condition y(0) = y_0

y(0) = -0 - sqrt( 4 - 0^2 )

= -sqrt(4)

= -2

C = -2

d/dt (y + t + sqrt( 4 - t^2 ))

d/dt (( 4 - t^2 )^(1/2) + y + t)

(1/2)(-2t)(4-t^2)^(-1/2) + 1

(-t)(4-t^2)^(-1/2) + 1

d/dy (y + t + sqrt( 4 - t^2 )) = 1

M = 1 - t(4-t^2)^(-1/2) + g(y) = 1 - t(4-t^2)^(-1/2) - 2 = - t(4-t^2)^(-1/2) -1

N = 1 + h(t)

h(t) = - t(4-t^2)^(-1/2) -1

(y + a t) y ' + (a y + b t) = 0

1 = (y + a t)

a = (1-y)/t

(a y + b t) = - t(4-t^2)^(-1/2) -1

((1-y)/t)y + bt = - t(4-t^2)^(-1/2) -1

bt = ((y-1)/t)y - t(4-t^2)^(-1/2) -1

b = ((y-1)/t^2)y - (4-t^2)^(-1/2) -1/t

Checking this solution:

Solution: (y + a t) y ' + (a y + b t) = 0

(y + ((1-y)/t)t)y' + (((1-y)/t)y + (((y-1)/t^2)y - (4-t^2)^(-1/2) -1/t)t = 0

(y + (1-y))y' + (((1-y)/t)y + (((y-1)/t)y - t(4-t^2)^(-1/2) -1) = 0

(1)y' - t(4-t^2)^(-1/2) -1 = 0

M = - t(4-t^2)^(-1/2) -1

N = 1

confidence rating #$&*:

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

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

Given Solution:

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

Self-critique (if necessary):

Self-critique rating:

Self-critique (if necessary):

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

Self-critique rating:

Query_06

#$&*

course MTH-279

06/21 around 5:56AM.

Query 06 Differential Equations*********************************************

Question: 3.3.4. If the equation is exact, solve the equation (6 t + y^3 ) y ' + 3 t^2 y = 0, with y(2) = 1.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

(6 t + y^3 ) y ' + 3 t^2 y = 0

Since in the form Mdt + Ndy = 0

M = 3 t^2 y

N = (6 t + y^3 )

dM/dy = 3t^2

dN/dt = 6

Since these two are not equal the equation is not exact.

confidence rating #$&*:

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

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

Given Solution:

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

Self-critique (if necessary):

Self-critique rating:

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

Question: If the equation is exact, solve the equation (6 t + 3 t^3 ) y ' + 6 y + 9/2 t^2 y^2 = -t, with y(2) = 1. *

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

(6 t + 3 t^3 ) y ' + 6 y + 9/2 t^2 y^2 + t = 0

Since in the form Mdt + Ndy = 0

M = 6 y + 9/2 t^2 y^2 + t

N = (6 t + 3 t^3 )

dM/dy = 9yt^2 + 6

dN/dt = 9t^2 + 6

These are not equivalent either, so the equation is not exact.

confidence rating #$&*:

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

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

Given Solution:

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

Self-critique (if necessary):

Self-critique rating:

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

Question: 3.3.6. If the equation is exact, solve the equation y ' = - ( y cos(t y) + 1) / (t cos(t y) + 2 y e^y^2) with y(0) = pi.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

y ' = ( y cos(t y) + 1) / (t cos(t y) + 2 y e^y^2)

(t cos(t y) + 2 y e^y^2)y ' = ( y cos(t y) + 1)

(t cos(t y) + 2 y e^y^2)y ' - ( y cos(t y) + 1) = 0

M = - ( y cos(t y) + 1) = -y cos(t y) - 1

N = (t cos(t y) + 2 y e^y^2)

dM/dy = tysin(ty)-cos(ty)

dN/dt = cos(ty) - tysin(ty)

These are close but not equivalent either.

@&
If the equation is exact, the N function is the y derivative of our unknown 'source' function F(x, y), and M is the t derivative of this function. It follows that we can find the F function by integration.

The differential of our unknown F function would be

F_y dy + F_t dt.

If it is so, the t derivative of F_y is equal to the y derivative of F_t, since F_yt = F_ty.

So the test for whether such an F function exists is whether the coefficient function of dy has a t derivative equal to the coefficient function of dt.

In terms of N and M, this means that N would be the y derivative of our F function and M the t derivative, so that N_t = M_y.

So:

Our equation

y ' = - ( y cos(t y) + 1) / (t cos(t y) + 2 y e^y^2)

rearranges to

(t cos(t y) + 2 y e^y^2) dy + ( y cos(t y) + 1) dt = 0.

This is exact if the t derivative of

(t cos(t y) + 2 y e^y^2)

is equal to the y derivative of

( y cos(t y) + 1) dt = 0.

Using the N dy + M dt notation of the text:

N = t cos(t y) + 2y e^(y^2)
and
M = y cos(t y) + 1
N_t = cos(t y) - y t sin(t y)
and
M_y = cos(t y) - y t sin(t y).

These are identical and the equation is indeed exact.

See if you can solve the equation. If you get the expression

-sin(t y) + e^(y^2)

at some point chances are you've just about got it. If not, you'd better ask for more review of your work.

You are of course always welcome to submit a revised solution using the Question Form.

*@

confidence rating #$&*:

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

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

Given Solution:

Self-critique (if necessary):

Self-critique rating:

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

Question: 3.3.10. If N(y, t) * y ' + t^2 + y^2 sin(t) = 0 is exact, what is the most general possible form of N(y, t)?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

N(y, t) * y ' + t^2 + y^2 sin(t) = 0

M = t^2 + y^2 sin(t)

N = ?

dM/dy = 2y sin(t)

Int(2y sin(t)) dt = -2ycos(t)

N = -2ycos(t)

@&
This isn't the most general form.

You don't appear to have considered the integration constant, which is the most general function of y and t whose derivative with respect to t is zero.
*@

Confidence rating: 2

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

Given Solution:

Self-critique (if necessary):

Self-critique rating:

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

Question: 3.3.12. If y = -t - sqrt( 4 - t^2 ) is the solution of the differential equation (y + a t) y ' + (a y + b t) = 0 with initial condition y(0) = y_0, what are the values of a, b and y_0?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

y = -t - sqrt( 4 - t^2 )

Solution: (y + a t) y ' + (a y + b t) = 0

First we can solve the condition y(0) = y_0

y(0) = -0 - sqrt( 4 - 0^2 )

= -sqrt(4)

= -2

C = -2

d/dt (y + t + sqrt( 4 - t^2 ))

d/dt (( 4 - t^2 )^(1/2) + y + t)

(1/2)(-2t)(4-t^2)^(-1/2) + 1

(-t)(4-t^2)^(-1/2) + 1

d/dy (y + t + sqrt( 4 - t^2 )) = 1

M = 1 - t(4-t^2)^(-1/2) + g(y) = 1 - t(4-t^2)^(-1/2) - 2 = - t(4-t^2)^(-1/2) -1

N = 1 + h(t)

h(t) = - t(4-t^2)^(-1/2) -1

(y + a t) y ' + (a y + b t) = 0

1 = (y + a t)

a = (1-y)/t

(a y + b t) = - t(4-t^2)^(-1/2) -1

((1-y)/t)y + bt = - t(4-t^2)^(-1/2) -1

bt = ((y-1)/t)y - t(4-t^2)^(-1/2) -1

b = ((y-1)/t^2)y - (4-t^2)^(-1/2) -1/t

@&
The equation is exact, with N_t = M_y = a.

a, b and y_0 are constants. Your solution gives a and b as nonconstant functions of y and t.
*@

Checking this solution:

Solution: (y + a t) y ' + (a y + b t) = 0

(y + ((1-y)/t)t)y' + (((1-y)/t)y + (((y-1)/t^2)y - (4-t^2)^(-1/2) -1/t)t = 0

(y + (1-y))y' + (((1-y)/t)y + (((y-1)/t)y - t(4-t^2)^(-1/2) -1) = 0

(1)y' - t(4-t^2)^(-1/2) -1 = 0

M = - t(4-t^2)^(-1/2) -1

N = 1

confidence rating #$&*:

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

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

Given Solution:

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

Self-critique (if necessary):

Self-critique rating:

Self-critique (if necessary):

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

Self-critique rating:

#*&!

Query_06

#$&*

course MTH-279

06/21 around 5:56AM.

Query 06 Differential Equations*********************************************

Question: 3.3.4. If the equation is exact, solve the equation (6 t + y^3 ) y ' + 3 t^2 y = 0, with y(2) = 1.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

(6 t + y^3 ) y ' + 3 t^2 y = 0

Since in the form Mdt + Ndy = 0

M = 3 t^2 y

N = (6 t + y^3 )

dM/dy = 3t^2

dN/dt = 6

Since these two are not equal the equation is not exact.

confidence rating #$&*:

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

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

Given Solution:

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

Self-critique (if necessary):

Self-critique rating:

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

Question: If the equation is exact, solve the equation (6 t + 3 t^3 ) y ' + 6 y + 9/2 t^2 y^2 = -t, with y(2) = 1. *

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

(6 t + 3 t^3 ) y ' + 6 y + 9/2 t^2 y^2 + t = 0

Since in the form Mdt + Ndy = 0

M = 6 y + 9/2 t^2 y^2 + t

N = (6 t + 3 t^3 )

dM/dy = 9yt^2 + 6

dN/dt = 9t^2 + 6

These are not equivalent either, so the equation is not exact.

confidence rating #$&*:

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

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

Given Solution:

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

Self-critique (if necessary):

Self-critique rating:

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

Question: 3.3.6. If the equation is exact, solve the equation y ' = - ( y cos(t y) + 1) / (t cos(t y) + 2 y e^y^2) with y(0) = pi.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

y ' = ( y cos(t y) + 1) / (t cos(t y) + 2 y e^y^2)

(t cos(t y) + 2 y e^y^2)y ' = ( y cos(t y) + 1)

(t cos(t y) + 2 y e^y^2)y ' - ( y cos(t y) + 1) = 0

M = - ( y cos(t y) + 1) = -y cos(t y) - 1

N = (t cos(t y) + 2 y e^y^2)

dM/dy = tysin(ty)-cos(ty)

dN/dt = cos(ty) - tysin(ty)

These are close but not equivalent either.

@&
If the equation is exact, the N function is the y derivative of our unknown 'source' function F(x, y), and M is the t derivative of this function. It follows that we can find the F function by integration.

The differential of our unknown F function would be

F_y dy + F_t dt.

If it is so, the t derivative of F_y is equal to the y derivative of F_t, since F_yt = F_ty.

So the test for whether such an F function exists is whether the coefficient function of dy has a t derivative equal to the coefficient function of dt.

In terms of N and M, this means that N would be the y derivative of our F function and M the t derivative, so that N_t = M_y.

So:

Our equation

y ' = - ( y cos(t y) + 1) / (t cos(t y) + 2 y e^y^2)

rearranges to

(t cos(t y) + 2 y e^y^2) dy + ( y cos(t y) + 1) dt = 0.

This is exact if the t derivative of

(t cos(t y) + 2 y e^y^2)

is equal to the y derivative of

( y cos(t y) + 1) dt = 0.

Using the N dy + M dt notation of the text:

N = t cos(t y) + 2y e^(y^2)
and
M = y cos(t y) + 1
N_t = cos(t y) - y t sin(t y)
and
M_y = cos(t y) - y t sin(t y).

These are identical and the equation is indeed exact.

See if you can solve the equation. If you get the expression

-sin(t y) + e^(y^2)

at some point chances are you've just about got it. If not, you'd better ask for more review of your work.

You are of course always welcome to submit a revised solution using the Question Form.

*@

confidence rating #$&*:

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

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

Given Solution:

Self-critique (if necessary):

Self-critique rating:

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

Question: 3.3.10. If N(y, t) * y ' + t^2 + y^2 sin(t) = 0 is exact, what is the most general possible form of N(y, t)?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

N(y, t) * y ' + t^2 + y^2 sin(t) = 0

M = t^2 + y^2 sin(t)

N = ?

dM/dy = 2y sin(t)

Int(2y sin(t)) dt = -2ycos(t)

N = -2ycos(t)

@&
This isn't the most general form.

You don't appear to have considered the integration constant, which is the most general function of y and t whose derivative with respect to t is zero.
*@

Confidence rating: 2

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

Given Solution:

Self-critique (if necessary):

Self-critique rating:

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

Question: 3.3.12. If y = -t - sqrt( 4 - t^2 ) is the solution of the differential equation (y + a t) y ' + (a y + b t) = 0 with initial condition y(0) = y_0, what are the values of a, b and y_0?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

y = -t - sqrt( 4 - t^2 )

Solution: (y + a t) y ' + (a y + b t) = 0

First we can solve the condition y(0) = y_0

y(0) = -0 - sqrt( 4 - 0^2 )

= -sqrt(4)

= -2

C = -2

d/dt (y + t + sqrt( 4 - t^2 ))

d/dt (( 4 - t^2 )^(1/2) + y + t)

(1/2)(-2t)(4-t^2)^(-1/2) + 1

(-t)(4-t^2)^(-1/2) + 1

d/dy (y + t + sqrt( 4 - t^2 )) = 1

M = 1 - t(4-t^2)^(-1/2) + g(y) = 1 - t(4-t^2)^(-1/2) - 2 = - t(4-t^2)^(-1/2) -1

N = 1 + h(t)

h(t) = - t(4-t^2)^(-1/2) -1

(y + a t) y ' + (a y + b t) = 0

1 = (y + a t)

a = (1-y)/t

(a y + b t) = - t(4-t^2)^(-1/2) -1

((1-y)/t)y + bt = - t(4-t^2)^(-1/2) -1

bt = ((y-1)/t)y - t(4-t^2)^(-1/2) -1

b = ((y-1)/t^2)y - (4-t^2)^(-1/2) -1/t

@&
The equation is exact, with N_t = M_y = a.

a, b and y_0 are constants. Your solution gives a and b as nonconstant functions of y and t.
*@

Checking this solution:

Solution: (y + a t) y ' + (a y + b t) = 0

(y + ((1-y)/t)t)y' + (((1-y)/t)y + (((y-1)/t^2)y - (4-t^2)^(-1/2) -1/t)t = 0

(y + (1-y))y' + (((1-y)/t)y + (((y-1)/t)y - t(4-t^2)^(-1/2) -1) = 0

(1)y' - t(4-t^2)^(-1/2) -1 = 0

M = - t(4-t^2)^(-1/2) -1

N = 1

confidence rating #$&*:

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

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

Given Solution:

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

Self-critique (if necessary):

Self-critique rating:

Self-critique (if necessary):

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

Self-critique rating:

#*&!#*&!

Query_06

#$&*

course MTH-279

06/21 around 5:56AM.

Query 06 Differential Equations*********************************************

Question: 3.3.4. If the equation is exact, solve the equation (6 t + y^3 ) y ' + 3 t^2 y = 0, with y(2) = 1.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

(6 t + y^3 ) y ' + 3 t^2 y = 0

Since in the form Mdt + Ndy = 0

M = 3 t^2 y

N = (6 t + y^3 )

dM/dy = 3t^2

dN/dt = 6

Since these two are not equal the equation is not exact.

confidence rating #$&*:

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

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

Given Solution:

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

Self-critique (if necessary):

Self-critique rating:

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

Question: If the equation is exact, solve the equation (6 t + 3 t^3 ) y ' + 6 y + 9/2 t^2 y^2 = -t, with y(2) = 1. *

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

(6 t + 3 t^3 ) y ' + 6 y + 9/2 t^2 y^2 + t = 0

Since in the form Mdt + Ndy = 0

M = 6 y + 9/2 t^2 y^2 + t

N = (6 t + 3 t^3 )

dM/dy = 9yt^2 + 6

dN/dt = 9t^2 + 6

These are not equivalent either, so the equation is not exact.

confidence rating #$&*:

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

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

Given Solution:

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

Self-critique (if necessary):

Self-critique rating:

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

Question: 3.3.6. If the equation is exact, solve the equation y ' = - ( y cos(t y) + 1) / (t cos(t y) + 2 y e^y^2) with y(0) = pi.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

y ' = ( y cos(t y) + 1) / (t cos(t y) + 2 y e^y^2)

(t cos(t y) + 2 y e^y^2)y ' = ( y cos(t y) + 1)

(t cos(t y) + 2 y e^y^2)y ' - ( y cos(t y) + 1) = 0

M = - ( y cos(t y) + 1) = -y cos(t y) - 1

N = (t cos(t y) + 2 y e^y^2)

dM/dy = tysin(ty)-cos(ty)

dN/dt = cos(ty) - tysin(ty)

These are close but not equivalent either.

@&
If the equation is exact, the N function is the y derivative of our unknown 'source' function F(x, y), and M is the t derivative of this function. It follows that we can find the F function by integration.

The differential of our unknown F function would be

F_y dy + F_t dt.

If it is so, the t derivative of F_y is equal to the y derivative of F_t, since F_yt = F_ty.

So the test for whether such an F function exists is whether the coefficient function of dy has a t derivative equal to the coefficient function of dt.

In terms of N and M, this means that N would be the y derivative of our F function and M the t derivative, so that N_t = M_y.

So:

Our equation

y ' = - ( y cos(t y) + 1) / (t cos(t y) + 2 y e^y^2)

rearranges to

(t cos(t y) + 2 y e^y^2) dy + ( y cos(t y) + 1) dt = 0.

This is exact if the t derivative of

(t cos(t y) + 2 y e^y^2)

is equal to the y derivative of

( y cos(t y) + 1) dt = 0.

Using the N dy + M dt notation of the text:

N = t cos(t y) + 2y e^(y^2)
and
M = y cos(t y) + 1
N_t = cos(t y) - y t sin(t y)
and
M_y = cos(t y) - y t sin(t y).

These are identical and the equation is indeed exact.

See if you can solve the equation. If you get the expression

-sin(t y) + e^(y^2)

at some point chances are you've just about got it. If not, you'd better ask for more review of your work.

You are of course always welcome to submit a revised solution using the Question Form.

*@

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Given Solution:

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Question: 3.3.10. If N(y, t) * y ' + t^2 + y^2 sin(t) = 0 is exact, what is the most general possible form of N(y, t)?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

N(y, t) * y ' + t^2 + y^2 sin(t) = 0

M = t^2 + y^2 sin(t)

N = ?

dM/dy = 2y sin(t)

Int(2y sin(t)) dt = -2ycos(t)

N = -2ycos(t)

@&
This isn't the most general form.

You don't appear to have considered the integration constant, which is the most general function of y and t whose derivative with respect to t is zero.
*@

Confidence rating: 2

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Given Solution:

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Question: 3.3.12. If y = -t - sqrt( 4 - t^2 ) is the solution of the differential equation (y + a t) y ' + (a y + b t) = 0 with initial condition y(0) = y_0, what are the values of a, b and y_0?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

y = -t - sqrt( 4 - t^2 )

Solution: (y + a t) y ' + (a y + b t) = 0

First we can solve the condition y(0) = y_0

y(0) = -0 - sqrt( 4 - 0^2 )

= -sqrt(4)

= -2

C = -2

d/dt (y + t + sqrt( 4 - t^2 ))

d/dt (( 4 - t^2 )^(1/2) + y + t)

(1/2)(-2t)(4-t^2)^(-1/2) + 1

(-t)(4-t^2)^(-1/2) + 1

d/dy (y + t + sqrt( 4 - t^2 )) = 1

M = 1 - t(4-t^2)^(-1/2) + g(y) = 1 - t(4-t^2)^(-1/2) - 2 = - t(4-t^2)^(-1/2) -1

N = 1 + h(t)

h(t) = - t(4-t^2)^(-1/2) -1

(y + a t) y ' + (a y + b t) = 0

1 = (y + a t)

a = (1-y)/t

(a y + b t) = - t(4-t^2)^(-1/2) -1

((1-y)/t)y + bt = - t(4-t^2)^(-1/2) -1

bt = ((y-1)/t)y - t(4-t^2)^(-1/2) -1

b = ((y-1)/t^2)y - (4-t^2)^(-1/2) -1/t

@&
The equation is exact, with N_t = M_y = a.

a, b and y_0 are constants. Your solution gives a and b as nonconstant functions of y and t.
*@

Checking this solution:

Solution: (y + a t) y ' + (a y + b t) = 0

(y + ((1-y)/t)t)y' + (((1-y)/t)y + (((y-1)/t^2)y - (4-t^2)^(-1/2) -1/t)t = 0

(y + (1-y))y' + (((1-y)/t)y + (((y-1)/t)y - t(4-t^2)^(-1/2) -1) = 0

(1)y' - t(4-t^2)^(-1/2) -1 = 0

M = - t(4-t^2)^(-1/2) -1

N = 1

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Given Solution:

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

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Self-critique rating:

#*&!#*&!#*&!

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

#*&!

#*&!#*&!

#*&!#*&!#*&!

Query_06

Query_06

@& This isn't the most general form. You don't appear to have considered the integration constant, which is the most general function of y and t whose derivative with respect to t is zero. *@

@& The equation is exact, with N_t = M_y = a. a, b and y_0 are constants. Your solution gives a and b as nonconstant functions of y and t. *@

Query_06

@& This isn't the most general form. You don't appear to have considered the integration constant, which is the most general function of y and t whose derivative with respect to t is zero. *@

@& The equation is exact, with N_t = M_y = a. a, b and y_0 are constants. Your solution gives a and b as nonconstant functions of y and t. *@

Query_06

@& This isn't the most general form. You don't appear to have considered the integration constant, which is the most general function of y and t whose derivative with respect to t is zero. *@

@& The equation is exact, with N_t = M_y = a. a, b and y_0 are constants. Your solution gives a and b as nonconstant functions of y and t. *@

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

@&
This isn't the most general form.

You don't appear to have considered the integration constant, which is the most general function of y and t whose derivative with respect to t is zero.
*@

@&
The equation is exact, with N_t = M_y = a.

a, b and y_0 are constants. Your solution gives a and b as nonconstant functions of y and t.
*@

@&
This isn't the most general form.

You don't appear to have considered the integration constant, which is the most general function of y and t whose derivative with respect to t is zero.
*@

@&
The equation is exact, with N_t = M_y = a.

a, b and y_0 are constants. Your solution gives a and b as nonconstant functions of y and t.
*@

@&
This isn't the most general form.

You don't appear to have considered the integration constant, which is the most general function of y and t whose derivative with respect to t is zero.
*@

@&
The equation is exact, with N_t = M_y = a.

a, b and y_0 are constants. Your solution gives a and b as nonconstant functions of y and t.
*@