Query 11

#$&*

course MTH 279

10:19 am 3/4 I've submitted two other queries that haven't been posted and a question form response also hasn't been posted and that's been several days ago now.

Query 11 Differential Equations*********************************************

Question:  3.8.4.  Solve the equation y ' = - y + t with y(0) = 0.

Write the expression y_(k + 1) = y_k + h f(t_k, y_k) for h = .01.

Find y_k for k = 1, 2, 3.

Using your original solution for the equation, compare your values of y_k with the values given by the accurate solution.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:  

 y = t+ e^(-t)-1

y_(k+1) = y_k + 0.01(-(y_k+t_k))

t_0=0, y_0= 0 --> y_0' = 0

t_1 = 0.01, y_1= 0.00, y_1' = 0

y_2 = 0.00 + 0.01*0.01= 0.0001

t_2= 0.02, y_2' = 0.0199

y_3 = 0.0001 + 0.01(0.0199) = 0.0003

y_1 = 0.0000

y_2 = 0.0001

y_3= 0.0003

Actual y_k

y_1 = 0.00005

y_2 = 0.00019

y_3 = 0.00045

confidence rating #$&*:

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

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

Given Solution: 

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

Self-critique (if necessary):

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

Self-critique rating:

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

Question:  3.8.6.  Euler's method applied to the equation y ' = alpha t + beta, y(t_0) = y_0, yields y values -1, -.9, -.81 and -.73 at respective t values 0, .1, .2, .3.  Find the values of alpha, beta, t_0 and y_0.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:  

Seems t_0 and y_0 are given:

t_0 = 0 --> y_0 = -1

(-0.9 - (-1))/(0.2-0.1)

y_1 = -0.9 = -1 + 0.1*y'(0.1)

y'= 1

y_2 = -0.81 = -0.9 + 0.1y'(0.2)

y'= 0.9

2 eqs 2 unknowns

y'(0.1) = 1= alpha*0.1+beta

y'(0.2)= 0.9 = alpha*0.2 + beta

alpha = -1

beta = 1.1

Check:

y(0.3) = -0.73= -0.81 + 0.1y'(0.3)

y' = 0.8

y'(0.3)= -(0.3) + 1.1 = 0.8

confidence rating #$&*:

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

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

Given Solution: 

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

Self-critique (if necessary):

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

Self-critique rating:

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

Question:  3.8.8a.  For each of the following situations, will Euler's method overestimate or underestimate the values of the solution to an equation:

• The solution curve is known to be increasing and concave up.

• The solution curve is known to be increasing and concave down.

• The solution curve is known to be decreasing and concave up.

• The solution curve is known to be decreasing and concave down.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution: 

Due to the geometry of the tangent line, since this is a tangent or linear approximation, the over and under are:

Inc&Up: under

Inc&Down: over

Dec&Up: under

Dec&Down: over

This is a linear approximation because the eq used to find the new y value is linear. The estimates are easily seen if the graphs are drawn along with tangent lines at certain points and if the tangent line goes above the curve it's overestimated and if it's under the line it's underestimated.

 

confidence rating #$&*:

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

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

Given Solution: 

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

Self-critique (if necessary):

 Self-critique rating:

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

Question:  3.8.14.  y ' = y^2 with y(0) = 1.

Solve the equation.

Perform Euler's Method to approximate the values of the solution on the t interval [0, 1.2] with step size h = .1.

Compare the values you get with the values given by your solution to the equation.

This could be done by hand, but it would take awhile and the probability of an error would be relatively high.  A spreadsheet is recommended.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution: 

The solution is:

y = 1/(1-t)

The approximations work up until t =1. The asymptote at t =1 is not predicted by the equation and isn't reflected in the y values since the t=1 iteration has a y value when the solution function is undefined. After the t= 1 iteration the y values are completely wrong since they're actually negative but the method gives positive values. I think this has to do with the fact that t isn't in our original equation. Also, as t--> 1 y grows very large very quickly but my values don't start getting large until t = 1.3 using the method, not to mention it should actually be negative then anyway. So it's a sever underestimate as t --> 1 and it will never give negative values for y since y' is always positive. I'm new to Excel so I may not have my formula setup right but it seems to work the way you showed us.

confidence rating #$&*:

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

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

Given Solution: 

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

Self-critique (if necessary):

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

Self-critique rating:

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

Question:  2. Solve the preceding question if the tank contains 500 gallons of 5% solution, and the goal is to achieve 1000 gallons of 3.5% solution at the end of 8 hours.  Assume that no solution is removed from the tank until it is full, and that once the tank is full, the resulting overflow is well-mixed.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution: 

I don't know why this question is here because this is from Ch 2 even though there is a tank question in this section this isn't it and it's not even close to the same question either.

confidence rating #$&*:

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

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

Given Solution: 

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

Self-critique (if necessary):

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

Self-critique rating:"

Self-critique (if necessary):

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

Self-critique rating:

&#Good work. Let me know if you have questions. &#