Query_06

#$&*

course Mth 279

3/19

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 2 are not equal the Equation is not Exact.

confidence rating #$&*:

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

2

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

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 eqivilant either, so the Equation is not Exact.

confidence rating #$&*:

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

2

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

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 Equivilant either.

@&

The problem in the text was in fact

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

So the 'close' of your solution, which is correct if the - sign isn't there, become 'is'.

We test the equation to see if it is exact, i.e., of the form

M dy + N dt = 0

with M_t = N_x.

The equation can be rearranged to

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

so it is of the form M dy + N dt = 0 with

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

and

N = y cos(t y) + 1

M_t = cos(t y) - y t sin(t y)

and

N_y = cos(t y) - y t sin(t y)

These are equal, so our equation is of the form

dF = 0

with

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

and

F_t = y cos(t y) + 1.

F_y = t cos(t y) + 2y e^(y^2) implies that

F = integral( (t cos(t y) + 2 y e^(y^2) ) dy = -sin(t y) + e^(y^2) + g(t),

where g(t) is our integration constant (y being the variable of integration, any function of t has derivative zero with respect to y and so is constant in this integral).

F_t = y cos(t y) + 1 implies that

F = integral( (y cos(t y) + 1) dt) = -sin(t y) + h(y),

where h(y) is constant with respect to t, the variable of integration.

If h(y) = e^(y^2) and g(t) = 0, our result is

F = -sin(t y) + e^(y^2) .

dF = 0 means that

F = c,

where c is constant, so

-sin(t y) + e^(y^2) = c.

The initial condition is that y(0) = pi, so we have

-sin(0) + e^(pi^2) = c

and c = e^(pi^2).

Our implicit solution is therefore given by the equation

-sin(t y) + e^(y^2) = e^(pi^2).

*@

confidence rating #$&*:

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

2

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

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:

This equation can be worked backwards.

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)

@&

N can be more general than that.

Specifically, adding any arbitrary function of y to your result we still end up with dN/dt = 2 y sin(t).

So the most general form is

N = - 2 y cos(t) + g(y),

where g(y) can be any function of y.

*@

confidence rating #$&*:

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

3

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

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:

This problem also is able to be worked backwards.

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

Solution checks out!

confidence rating #$&*:

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

3

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

Given Solution:

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

Self-critique (if necessary):

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

Self-critique rating:"

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