Query_19

#$&*

course MTH-279

07/29 around 11:20PMJust turning in some work I did leading up to test two.

Query 19 Differential Equations

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Question: Find the general solution of the equation

y '' + y = e^t sin(t).

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

X^2 = -1

sqrtx = sqrt -1

x = +/- i

---

lambda1 = i

lambda2 = -i

----

y = C1 e^(lambda1 t) + C2 e^(lambda2 t)

y = C1 e^(i t) + C2 e^ (-i t)

y = C1 cost + C2sint

----

yp(t) = Aetsint + Be^tcost

y’p(t) = Ae^tcost + Ae^tsint -Be^tsint + Be^tcost

y’’p(t) = -(A+B)e^t sint + (A+B)e^t cost + (A-B)e^t cost + (A-B)e^t sint

-----

-2Be^t sint + 2A e^t cost + Be^t cost + Ae^t sint = e^t sint

----

-2B + A = 1

2A + B = 0

---

-2B + A = 1

4A + 2B = 0

----

5A = 1 -> A = 1/5

---

2A + B = 0

2(1/5) + B = 0

-2/5 = B

----

y(t) = C1 cost + C2 sint + 1/5 e^t sint - 2/5 e^t cost

Confidence rating: 3

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Given Solution:

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

Self-critique (if necessary): OK

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Self-critique rating: OK

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Question: Find the general solution of the equation

y '' + y ' = 6 t^2

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

x^2 + x = 0

x(x+1) = 0

----

x = 0, x = -1

lambda1 = 0

lambda2 = -1

---

y = C1 e^(lambda1 t) + C2e^(lambda2 t)

y = C1 e^((0)t) + C2 e ^((-1)t)

---

y = C1 + C2 e^-t

---

to get t^2 on left side, need t^2 coefficient

----

yp(t) = At^2 + Bt + C

y’p(t) = 2At + B

y’’p(t) = 2A -- no t^2 in this set, so we need to move to a higher power

----

yp(t) = At^3 + Bt^2 + Ct + D

yp’(t) = 3At^2 + 2Bt + C

y’’p(t) = 6At + 2B -- use the last two in equation

----

y’’ + y’ = 6t^2

---

(6At+2B)+(3At^2 + 2Bt + C) = 6t^2 (match value of coefficient on each side of equation)

----

3A = 6 -> A = 2

---

6A + 2B = 0

12 + 2B = 0 -> B = -6

----

2B + C = 0

-12 + C = 0 -> C=12

----

yp(t) = 2t^3 - 6t^2 + 12t + D

---

recall y(t) = (yc(t)) + (yp(t)

y(t) = (C1 + C2e^-t) + (2t^3 - 6t^2 + 12t)

Confidence rating: 3

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Given Solution:

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

Self-critique (if necessary): OK

Self-critique rating: OK

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Question: Find the general solution of the equation

y '' + y ' = cos(t).

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

x^2 + x = 0

x(x+1) = 0

---

x = 0

x=-1

---

lambda1 = 0

lambda2 = -1

----

y = C1 e^(lambda1 t) + C2 e^(lambda2 t)

y = C1 e^((0)t) + C2 e^((-1)t)

y = C1 + C2 e^-t

----

yp(t) = Acost + Bsint

yp’(t) = -Asint + Bcost

yp’’(t) = -Acost -Bsint

----

-Acost -Bsint - Asint + Bcost = cost

---

-A + B = -1

-B - A = 0

----

-2A = 1 -> A= -1/2

-----

-(-1/2) + B = 1

½ + B + 1

B = ½

----

y(t) = C1 + C2 e^-t - ½ cost + 1/2sint

confidence rating #$&*:

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

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Given Solution:

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

Self-critique (if necessary): OK

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Self-critique rating: OK

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