Query 13

#$&*

course Mth 279

May 7 around 1:30pm. Resubmitting this, never showed up in portfolio.

Query 13 Differential Equations*********************************************

Question: Find the largest interval on which the equation

y '' + y ' + 3 t y = tan(t)

has a solution, with the initial conditions y(pi) = 1 and y ' (pi) = -1.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

y '' + y ' + 3 t y = tan(t)

(t0) = pi

There are no “restrictions” on this one, bc there is no denominators or square roots.

Theorem 4.1 says to let p(t), q(t), and g(t) be continuous functions on the interval (a,b) and let t0 be in (a,b), then the initial problem has a unique solution defined on the entire interval (a,b).

So, y0 = 1 and y0’ = (-1)

p(t) = 1

q(t) = 3t

g(t) = tan(t)

t-interval: (-inf, inf)

@& tan(t) is undefined for t = -3 pi / 2, - pi / 2, pi / 2, 3 pi / 2, ... .

pi lies in the interval (pi/2, 3 pi/2) so the given solution, with its initial conditions at t = pi, is guaranteed to exist in this interval.*@

confidence rating #$&*:

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

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

Given Solution:

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

Self-critique (if necessary):

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

Self-critique rating:

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

Question: Find the largest interval on which the equation

t y '' + sin(2 t) / (t^2 - 9) y ' + 2 y = 0

has a solution, with the initial conditions y(1) = 0, y ' (1) = 1.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

t y '' + sin(2 t) / (t^2 - 9) y ' + 2 y = 0

(t0) = 1

There ARE restrictions, due to the denominator, t CANNOT equal 3.

Theorem 4.1 says to let p(t), q(t), and g(t) be continuous functions on the interval (a,b) and let t0 be in (a,b), then the initial problem has a unique solution defined on the entire interval (a,b).

So, y0 = 0 and y0’ = 1

p(t) = sin(2 t) / (t^2 - 9)

q(t) = 2

g(t) = 0

t-interval: (-inf < t < 3)

@& When put into standard form

y '' + sin(2 t) / (t * (t^2 - 9) ) + 2 y / t = 0

we find that the equation is undefined for t = -3, 0 and 3.

The initial condition is given at t = 1, so the solution is guaranteed on the interval (0, 3).*@

confidence rating #$&*:

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

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

Given Solution:

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

Self-critique (if necessary):

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

Self-critique rating:

"

Self-critique (if necessary):

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

Self-critique rating:

________________________________________

#$&*

@& Right idea, but check my notes for some details you missed (e.g., the tan function isn't defined for all t, and the equation needs to be in standard form with y '' having constant coefficient, preferably 1).*@

Query 13

#$&*

course Mth 279

May 7 around 1:30pm. Resubmitting this, never showed up in portfolio.

Query 13 Differential Equations*********************************************

Question: Find the largest interval on which the equation

y '' + y ' + 3 t y = tan(t)

has a solution, with the initial conditions y(pi) = 1 and y ' (pi) = -1.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

y '' + y ' + 3 t y = tan(t)

(t0) = pi

There are no “restrictions” on this one, bc there is no denominators or square roots.

Theorem 4.1 says to let p(t), q(t), and g(t) be continuous functions on the interval (a,b) and let t0 be in (a,b), then the initial problem has a unique solution defined on the entire interval (a,b).

So, y0 = 1 and y0’ = (-1)

p(t) = 1

q(t) = 3t

g(t) = tan(t)

t-interval: (-inf, inf)

@& tan(t) is undefined for t = -3 pi / 2, - pi / 2, pi / 2, 3 pi / 2, ... .

pi lies in the interval (pi/2, 3 pi/2) so the given solution, with its initial conditions at t = pi, is guaranteed to exist in this interval.*@

confidence rating #$&*:

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

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

Given Solution:

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

Self-critique (if necessary):

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

Self-critique rating:

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

Question: Find the largest interval on which the equation

t y '' + sin(2 t) / (t^2 - 9) y ' + 2 y = 0

has a solution, with the initial conditions y(1) = 0, y ' (1) = 1.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

t y '' + sin(2 t) / (t^2 - 9) y ' + 2 y = 0

(t0) = 1

There ARE restrictions, due to the denominator, t CANNOT equal 3.

Theorem 4.1 says to let p(t), q(t), and g(t) be continuous functions on the interval (a,b) and let t0 be in (a,b), then the initial problem has a unique solution defined on the entire interval (a,b).

So, y0 = 0 and y0’ = 1

p(t) = sin(2 t) / (t^2 - 9)

q(t) = 2

g(t) = 0

t-interval: (-inf < t < 3)

@& When put into standard form

y '' + sin(2 t) / (t * (t^2 - 9) ) + 2 y / t = 0

we find that the equation is undefined for t = -3, 0 and 3.

The initial condition is given at t = 1, so the solution is guaranteed on the interval (0, 3).*@

confidence rating #$&*:

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

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

Given Solution:

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

Self-critique (if necessary):

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

Self-critique rating:

"

Self-critique (if necessary):

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

Self-critique rating:

________________________________________

#$&*

@& Right idea, but check my notes for some details you missed (e.g., the tan function isn't defined for all t, and the equation needs to be in standard form with y '' having constant coefficient, preferably 1).*@