Query_17

#$&*

course MTH-279

07/29 around 10:50PM.Just turning in some problems I did leading up to test two.

Query 17 Differential Equations

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Question: Solve the equation

25 y '' + 20 y ' + 4 y = 0, y(5) = 4 e^-2, y ' (5) = -3/5 e^-2

with y(5) = 4 e^-2 and y ' (5) = -3/5 e^-2.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Divide 25 to get y’’ by itself

y’’ + 20/25y’ + 4y/25 = 0

x^2 + 20/25x + 4/25 = 0

x^2 + 4/5x + 4/25 = 0

using the quadratic equation:

x = -4/5 +/- sqrt (4/5)^2 - 4(1)(4/25) / 2(1)

x = -4/5 +/- sqrt 0 / 2

x = -4/5 / 2 -> x = -4/10 -> x = -2/5

lambda one = lambda two = -2/5 (repeated real root)

so, y(t) = C1 e^ lambda t + C2 t e^lambda t

y(t) = C1 e^(-2/5)t + C2 t e^(-2/5)t

4e^-2 = C1e^(-2/5)(5) + C2 (5) e^(-2/5)(5)

4e^-2 = C1 e^-2 + C2 5 e^-2

4 = C1 + 5C2 - eq 1

----

y’(t) = -2/5 C1e^(-2/5)t + C2e^(-2/5)t - 2/5 t e^(-2/5)t [product rule]

-3/5e^-2 = -2/5C1e^(-2/5)(5) + C2 e^(-2/5)(5) - 2/5 C2 (5) e^(-2/5)(5)

-3/5e^-2 = -2/5C1e^-2 + C2e^-2 - 2C2e^-2

-3/5 = -2/5C1 - C2 - eq2

----

4 = C1 + 5C2

-3/5 = -2/5C1 - C2

----

-3 = -2C1 -5C2 (multiply by 5 to solve for C2)

+4 = C1 + 5C2

---

1 = -1C1

1 = -C1

C1 = -1

----

4 = -1 + 5C2

5 = 5C2

C2 = 1

Confidence rating: 3

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

Given Solution:

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

Self-critique (if necessary): OK

Self-critique rating: OK

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

Question: Solve the equation

3 y '' + 2 sqrt(3) y ' + y = 0, y(0) = 2 sqrt(3), y ' (0) = 3

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

divide by 3 to get y’’ by itself

y’’ + 2sqrt3y’ / 3 + y/3 = 0

x^2 + 2sqrt3 x / 3 + 1/3 = 0

using the quadratic equation:

x = -2sqrt3 /3 +/- sqrt (-2sqrt3 / 3)^2 - 4(1)(1/3) / 2(1)

x = -2sqrt3 +/- sqrt 12 - 12 / 6

x = -2sqrt3 / 6 -> x = -1sqrt3 / 3 (repeated real root)

so, y(t) = C1 e^ lambda t + C2 t e^lambda t

y(t) = C1 e^(-1sqrt3 / 3)t + C2 t e^(-1sqrt3 / 3)t

2sqrt3 = C1 e^(-sqrt3 / 3)(0) + C2 (0) e^(-sqrt3/3)(0)

2sqrt3 = C1 + 0

C1 = 2sqrt 3

-----

y’(t) = -sqrt3 / 3 C1 e^(-sqrt3 / 3)t - sqrt3 / 3 C2 t e^(-sqrt3 /3)t + C2 e (-sqrt3 /3)t

3 = -sqrt3/3 * 2sqrt3 e^(-sqrt3/3)(0) - sqrt3 /3 C2(0) e^(-sqrt3 /3)(0) + C2 e^(-sqrt3 / 3)(0)

3 = -2 + C2

5 = C2

confidence rating #$&*:

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

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

Given Solution:

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

Self-critique (if necessary): OK

Self-critique rating: OK

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

Question: Solve the equation

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

which has known solution y_1(t) = sin(t)

You will use reduction of order, find intervals of definition and interval(s) where the Wronskian is continuous and nonzero. See your text for a more complete statement of this problem.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Confidence rating:

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

Given Solution:

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

Self-critique (if necessary):

Self-critique rating:

Self-critique (if necessary):

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

Self-critique rating:

Query_17

#$&*

course MTH-279

07/29 around 10:50PM.Just turning in some problems I did leading up to test two.

Query 17 Differential Equations

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

Question: Solve the equation

25 y '' + 20 y ' + 4 y = 0, y(5) = 4 e^-2, y ' (5) = -3/5 e^-2

with y(5) = 4 e^-2 and y ' (5) = -3/5 e^-2.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Divide 25 to get y’’ by itself

y’’ + 20/25y’ + 4y/25 = 0

x^2 + 20/25x + 4/25 = 0

x^2 + 4/5x + 4/25 = 0

using the quadratic equation:

x = -4/5 +/- sqrt (4/5)^2 - 4(1)(4/25) / 2(1)

x = -4/5 +/- sqrt 0 / 2

x = -4/5 / 2 -> x = -4/10 -> x = -2/5

lambda one = lambda two = -2/5 (repeated real root)

so, y(t) = C1 e^ lambda t + C2 t e^lambda t

y(t) = C1 e^(-2/5)t + C2 t e^(-2/5)t

4e^-2 = C1e^(-2/5)(5) + C2 (5) e^(-2/5)(5)

4e^-2 = C1 e^-2 + C2 5 e^-2

4 = C1 + 5C2 - eq 1

----

y’(t) = -2/5 C1e^(-2/5)t + C2e^(-2/5)t - 2/5 t e^(-2/5)t [product rule]

-3/5e^-2 = -2/5C1e^(-2/5)(5) + C2 e^(-2/5)(5) - 2/5 C2 (5) e^(-2/5)(5)

-3/5e^-2 = -2/5C1e^-2 + C2e^-2 - 2C2e^-2

-3/5 = -2/5C1 - C2 - eq2

----

4 = C1 + 5C2

-3/5 = -2/5C1 - C2

----

-3 = -2C1 -5C2 (multiply by 5 to solve for C2)

+4 = C1 + 5C2

---

1 = -1C1

1 = -C1

C1 = -1

----

4 = -1 + 5C2

5 = 5C2

C2 = 1

Confidence rating: 3

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

Given Solution:

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

Self-critique (if necessary): OK

Self-critique rating: OK

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

Question: Solve the equation

3 y '' + 2 sqrt(3) y ' + y = 0, y(0) = 2 sqrt(3), y ' (0) = 3

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

divide by 3 to get y’’ by itself

y’’ + 2sqrt3y’ / 3 + y/3 = 0

x^2 + 2sqrt3 x / 3 + 1/3 = 0

using the quadratic equation:

x = -2sqrt3 /3 +/- sqrt (-2sqrt3 / 3)^2 - 4(1)(1/3) / 2(1)

x = -2sqrt3 +/- sqrt 12 - 12 / 6

x = -2sqrt3 / 6 -> x = -1sqrt3 / 3 (repeated real root)

so, y(t) = C1 e^ lambda t + C2 t e^lambda t

y(t) = C1 e^(-1sqrt3 / 3)t + C2 t e^(-1sqrt3 / 3)t

2sqrt3 = C1 e^(-sqrt3 / 3)(0) + C2 (0) e^(-sqrt3/3)(0)

2sqrt3 = C1 + 0

C1 = 2sqrt 3

-----

y’(t) = -sqrt3 / 3 C1 e^(-sqrt3 / 3)t - sqrt3 / 3 C2 t e^(-sqrt3 /3)t + C2 e (-sqrt3 /3)t

3 = -sqrt3/3 * 2sqrt3 e^(-sqrt3/3)(0) - sqrt3 /3 C2(0) e^(-sqrt3 /3)(0) + C2 e^(-sqrt3 / 3)(0)

3 = -2 + C2

5 = C2

confidence rating #$&*:

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

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

Given Solution:

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

Self-critique (if necessary): OK

Self-critique rating: OK

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

Question: Solve the equation

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

which has known solution y_1(t) = sin(t)

You will use reduction of order, find intervals of definition and interval(s) where the Wronskian is continuous and nonzero. See your text for a more complete statement of this problem.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Confidence rating:

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

Given Solution:

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

Self-critique (if necessary):

Self-critique rating:

Self-critique (if necessary):

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

Self-critique rating:

#*&!

Query_17

#$&*

course MTH-279

07/29 around 10:50PM.Just turning in some problems I did leading up to test two.

Query 17 Differential Equations

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

Question: Solve the equation

25 y '' + 20 y ' + 4 y = 0, y(5) = 4 e^-2, y ' (5) = -3/5 e^-2

with y(5) = 4 e^-2 and y ' (5) = -3/5 e^-2.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Divide 25 to get y’’ by itself

y’’ + 20/25y’ + 4y/25 = 0

x^2 + 20/25x + 4/25 = 0

x^2 + 4/5x + 4/25 = 0

using the quadratic equation:

x = -4/5 +/- sqrt (4/5)^2 - 4(1)(4/25) / 2(1)

x = -4/5 +/- sqrt 0 / 2

x = -4/5 / 2 -> x = -4/10 -> x = -2/5

lambda one = lambda two = -2/5 (repeated real root)

so, y(t) = C1 e^ lambda t + C2 t e^lambda t

y(t) = C1 e^(-2/5)t + C2 t e^(-2/5)t

4e^-2 = C1e^(-2/5)(5) + C2 (5) e^(-2/5)(5)

4e^-2 = C1 e^-2 + C2 5 e^-2

4 = C1 + 5C2 - eq 1

----

y’(t) = -2/5 C1e^(-2/5)t + C2e^(-2/5)t - 2/5 t e^(-2/5)t [product rule]

-3/5e^-2 = -2/5C1e^(-2/5)(5) + C2 e^(-2/5)(5) - 2/5 C2 (5) e^(-2/5)(5)

-3/5e^-2 = -2/5C1e^-2 + C2e^-2 - 2C2e^-2

-3/5 = -2/5C1 - C2 - eq2

----

4 = C1 + 5C2

-3/5 = -2/5C1 - C2

----

-3 = -2C1 -5C2 (multiply by 5 to solve for C2)

+4 = C1 + 5C2

---

1 = -1C1

1 = -C1

C1 = -1

----

4 = -1 + 5C2

5 = 5C2

C2 = 1

Confidence rating: 3

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

Given Solution:

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

Self-critique (if necessary): OK

Self-critique rating: OK

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

Question: Solve the equation

3 y '' + 2 sqrt(3) y ' + y = 0, y(0) = 2 sqrt(3), y ' (0) = 3

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

divide by 3 to get y’’ by itself

y’’ + 2sqrt3y’ / 3 + y/3 = 0

x^2 + 2sqrt3 x / 3 + 1/3 = 0

using the quadratic equation:

x = -2sqrt3 /3 +/- sqrt (-2sqrt3 / 3)^2 - 4(1)(1/3) / 2(1)

x = -2sqrt3 +/- sqrt 12 - 12 / 6

x = -2sqrt3 / 6 -> x = -1sqrt3 / 3 (repeated real root)

so, y(t) = C1 e^ lambda t + C2 t e^lambda t

y(t) = C1 e^(-1sqrt3 / 3)t + C2 t e^(-1sqrt3 / 3)t

2sqrt3 = C1 e^(-sqrt3 / 3)(0) + C2 (0) e^(-sqrt3/3)(0)

2sqrt3 = C1 + 0

C1 = 2sqrt 3

-----

y’(t) = -sqrt3 / 3 C1 e^(-sqrt3 / 3)t - sqrt3 / 3 C2 t e^(-sqrt3 /3)t + C2 e (-sqrt3 /3)t

3 = -sqrt3/3 * 2sqrt3 e^(-sqrt3/3)(0) - sqrt3 /3 C2(0) e^(-sqrt3 /3)(0) + C2 e^(-sqrt3 / 3)(0)

3 = -2 + C2

5 = C2

confidence rating #$&*:

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

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

Given Solution:

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

Self-critique (if necessary): OK

Self-critique rating: OK

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

Question: Solve the equation

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

which has known solution y_1(t) = sin(t)

You will use reduction of order, find intervals of definition and interval(s) where the Wronskian is continuous and nonzero. See your text for a more complete statement of this problem.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Confidence rating:

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

Given Solution:

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

Self-critique (if necessary):

Self-critique rating:

Self-critique (if necessary):

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

Self-critique rating:

#*&!#*&!

Query_17

Query_17

Query_17

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