Assignment 15

#$&*

course Mth 279

7/22

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’ + 3ty = tan(t)

so p(t) = 1 ; q(t) = 3t and g(t) = tan(t)

p and q are continuous everywhere

g discontinuous at every odd multiple of pi/2

t_0 = pi

so now we need to find the interval where t_0 falls in-between.

we would have intervals of (-3pi/2, -pi/2) , (-pi/2, pi/2), (pi/2, 3pi/2) and so on.

But the only integral that t_0 falls between is

(pi/2, 3pi/2)

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:

ty’’ + sin(2t)/(t^2 - 9) y’ + 2y = 0

divide by t

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

p = sin(2t)/(t(t^2 - 9)); q = 2y/t and g = 0

to find where p and q are discontinuous you find where the denominators are = 0

for p:

t(t^2 - 9) = 0

t = 0; t = 3, and t = -3

for q:

t= 0

so find the largest interval when t = 1

we would have the intervals of:

(-infinity, -3) (-3,0) (0,3) (3,infinity)

so t=1 falls between (0,3)

confidence rating #$&*:

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

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

Given Solution:

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

Self-critique (if necessary):

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

Self-critique rating:

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

Question: Tell whether each of the following is increasing or decreasing, and whether concave down or concave up, in the vicinity of the initial point:

y '' + y = 2 - sin(t), y(0) = 1, y ' (0) = -1.

y '' + y = - 2 t, y(0) = 1, y ' (0) = -1.

y '' - y = t^2, y(0) = 1, y ' (0) = 1.

y '' - y = - 2 cos(t), y(0) = 1, y ' (0) = 1.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

A) y’’ + y = 2 - sin(t)

slope = -1 so decreasing

y’’ = -1 + 2 -sin(0) = 1 so concave up

B) y’’ + y = -2t

initial point (0,1) with a slope of -1, so function is decreasing.

y’’ = -1 - 2(0) = -1 with a neg second derivative the function is concave down

C) y’’ -y = t^2

slope = 1 so function is increasing

y’’ = y + t^2 = 1 + 0^2 = 1 so function is concave up

D) y’’ - y = -2cos(t)

slope =1 so increasing

y’’ = 1 - 2cos(0) = 1 - 2(1) = -1 so concave down

confidence rating #$&*:

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

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

Given Solution:

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

Self-critique (if necessary):

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

Self-critique rating:"