Query 14

#$&*

course Mth 279

7/21 10:39 pm

Query 14 Differential Equations*********************************************

Question:  Decide whether y_1 = 3 e^t, y_2 = e^(t + 3) are solutions to the equation y '' - y = 0.  If so determine whether the two solutions are linearly independent.  If the solutions are linearly independent then find the general solution, as well as a particular solution for which y (-1) = 1 and y ' (-1) = 0.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution: 

 y_1 = 3e^t

y_1' = 3e^t

y_1'' = 3e^t

Plugging into equation

3e^t - 3e^t = 0

0 = 0

y_1 is a solution

y_2 = e^(t+3)

y_2' = e^(t+3)

y_2'' = e^(t+3)

Plugging in

e^(t+3) - e^(t+3) = 0

0 = 0

y_2 is a solution

Calculating Wronskian:

W = [ 3e^t e^(t+3),

3e^t e^(t+3) ]

 3e^(2t+3) - 3e^(2t+3) = 0

Because the Wronskian = 0, the solutions are not linearly independent.

 

 

confidence rating #$&*:

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

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

Given Solution: 

 

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

Self-critique (if necessary):

 

 

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

Self-critique rating:

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

Question:  Decide whether y_1 = e^(-t) and y_2 = 2 e^(1 - t) are solutions to the equation y '' + 2 y ' + y = 0.  If so determine whether the two solutions are linearly independent.  If the solutions are linearly independent then find the general solution, as well as a particular solution for which y (0) = 1 and y ' (0) = 0.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution: 

 

 y_1' = -e^(-t)

y_1'' = e^(-t)

Plugging in

e^(-t) - 2e^(-t) + e^(-t) = 0

0 = 0

y_1 is a solution

y_2' = -2e^(1-t)

y_2'' = 2e^(1-t)

Plugging in

2e^(1-t) - 4 e^(1-t) + 2e^(1-t) = 0

0 = 0

y_2 is a solution

Wronskian:

W = [e^(-t) 2e^(1-t),

-e^(-t) -2e^(1-t)]

W = -2e^(1-2t) + 2e^(1-2t) = 0

These solutions are not linearly independent.

 

 

confidence rating #$&*:

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

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

Given Solution: 

 

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

Self-critique (if necessary):

 

 

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

Self-critique rating:

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

Question:  Suppose y_1 and y_2 are solutions to the equation

y '' + alpha y ' + beta y = 0

and that y_1 = e^(2 t).  Suppose also that the Wronskian is e^(-t).

What are the values of alpha and beta?

 

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution: 

 

 W = [e^(2t) e^(At),

2e^(2t) Ae^(At)]

Ae^(2t + At) - 2Ae^(2t+At)

If W = e^(-t)

2+A = -1

A = -3

y_2 = e^(-3t)

We know that the equation is zero at -3 and 2

(y+3)(y-2)

y^2 +y -6

Alpha = 1

Beta = -6 

 

 

confidence rating #$&*:

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

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

Given Solution: 

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

Self-critique (if necessary):

 

 

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

Self-critique rating:

&#This looks very good. Let me know if you have any questions. &#