Assignment 13

#$&*

course Mth 279

7/8

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 = 0, 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:

A) y’ = -y+t

y’ +y = t

p(t) = 1

y(t) = Ce^-t + e^-t * int(e^t)*t dt

y(t) = Ce^-t + e^-t(te^t - e^t)

y(t) = Ce^-t + t -1

y(0) = Ce^0 + 0 -1

C =1

y(t) = e^-t + t - 1

B) y_(k + 1) = y_k + h (-y_k + t_k)

t_0 = 0 y_0 = 0

C)

when k = 0

y_1 = 0 + (-0.01 * 0) + 0.1(0)

y_1 = 0

when k = 1

y_2 = 0.01 + .01(-.01+0.2)

y_2 = 0.01

y_3 = .01+.01(-.01+.2)

y_3 = 0.029

y_4 = 0.029 + 0.01(-.029+ 0.3)

y_4 = 0.0561

D)

y_0 = e^(0) + 0 -1

y_0 = 0

y_1= e^(-.1) + .1 - 1

y_1 = 0.005

y_2 = e^(-.2) + .2 - 1

y_2 = 0.0187

y_3 = e^(-.3) + .3 - 1

y_3 = 0.041

@&

You haven't indicated the values of y(t) for the values of t corresponding to k = 0, 1, 2, 3.

You appear to have used increment h = 0.1 rather than .01.

You are doing most of the right things, but you need to use the right quantities in order to compare the Euler approximations with the actual values of the solution function.

*@

confidence rating #$&*:

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

1

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

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:

y’ = at + b

t_0 = 0

y_0 = -1

y1 = y0 + 0.1(at +b)

-0.9 = -1 + 0.1(a(0) +b)

0.1 =0.1b

b=1

-.81 = -.9 + 0.1(a(.1) + 1)

0.09 = 0.01a + .1

-0.01 = 0.01a

a= -1

confidence rating #$&*:

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

2

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

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:

1. under; tangent line is under solution curve

2. over; tangent line is over solution curve

3. under; tangent line is under solution curve

4. over; tangent line is over solution curve.

@&

Good.

*@

confidence rating #$&*:

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

1

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

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:

A) y’ = y^2

y^-2 dy = 1 dt

integrate

-y^-1 = t+C

-(1)^-1 = 0 + C

C = -1

y = 1/ (-t+1)

B.

when k=0

y1 = y0 + .1(y^2_0)

y_1 = 1.1

when k = 1

y_2 = 1.1 + .1(1.1)^2

y_2 = 1.221

when k=2

y_3 = 1.221 + .1(1.221)^2

y_3 = 1.370

and so on for a total of 12 times

D. Actual Values

y_1 = 1/(-.1-1) = 1.11

y_2 = 1/(-.2-1) = 1.25

y_3 = 1/(-.3-1) = 1.429

confidence rating #$&*:

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

2

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

Given Solution:

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

Self-critique (if necessary):

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

Self-critique rating:"

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