Query_02

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course MTH-279

06/15 around 8:30PM.

Solve each equation:*********************************************

Question: 1. y ' + y = 3

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Your solution:

w = y - 3

w' = y'

y = w + 3

y ' + y = 3

w' + w + 3 = 3

w' + w = 0

dw/dt + w = 0

dw/w = -1 dt

int (dw/w) = -1 int (dt)

ln(w) = -t + c

w = e^(-t + c)

w = c * e^(-t)

y-3 = c * e^(-t)

y = c * e^(-t) + 3

Confidence rating: 2

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Given Solution:

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Self-critique (if necessary):

Self-critique rating:

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Question:

2. y ' + t y = 3 t

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Your solution:

y = w + 3

y' = w'

w ' + t (w + 3) = 3 t

w ' + tw + 3t = 3 t

w ' + tw = 0

y' + p(t)y = 0 -> y = c*e^(-int(p(t)dt))

w = c*e^(-int(t dt)) = c*e^(-t^2/2)

y - 3 = c*e^(-t^2/2)

y = c*e^(-t^2/2) + 3

confidence rating #$&*:

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

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Given Solution:

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Self-critique (if necessary):

Self-critique rating:

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Question:

3. y ' - 4 y = sin(2 t)

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Your solution:

u(t) = e^(-int (4 dt))

y'e^(-int (4 dt)) - 4 y e^(-int (4 dt)) = sin(2 t)e^(-int (4 dt))

(e^(-int (4 dt)) * y)' = sin(2 t)e^(-int (4 dt))

int ((e^(-int (4 dt)) * y)') = int (sin(2 t)e^(-int (4 dt)))

e^(-int (4 dt)) * y = c - (2sin(2t) + cos(2t))/10e^(4t)

y = c*e^(4t) - (1/5)sin(2t) - (1/10)cos(2t)

confidence rating #$&*:

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

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Given Solution:

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Self-critique (if necessary):

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Self-critique rating:

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Question:

4. y ' + y = e^t, y (0) = 2

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Your solution:

u(t) = e^(int (1 dt)) = e^t

y' e^t + ye^t = e^(2t)

(ye^t)' = e^(2t)

int ((ye^t)') = int (e^(2t))

ye^t = e^(2t)/2 + c

y = e^t/2 + c/e^t

y(0) = 2

y(0) = e^0/2 + c/e^0 = 2

y(0) = 1/2 + c = 2

c = 3/2

y(t) = (1/2)e^t + (3/2)e^(-t)

confidence rating #$&*:

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

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Given Solution:

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Self-critique (if necessary):

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Self-critique rating:

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Question:

5. y ' + 3 y = 3 + 2 t + e^t, y(1) = e^2

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Your solution:

u(t) = e^(int (3 dt)) = e^(3t)

y ' + 3 y = 3 + 2 t + e^t

y 'e^(3t) + 3 ye^(3t) = 3e^(3t) + 2 te^(3t) + e^(4t)

(ye^(3t))' = 3e^(3t) + 2 te^(3t) + e^(4t)

int ((ye^(3t))') = int (3e^(3t) + 2 te^(3t) + e^(4t))

ye^(3t) = e^(3t) + (2/9) e^(3t) (3t - 1) + (1/4)e^(4t) + c

y(t) = c e^(-3t) + (2/3)t + (1/4)e^t + (7/9)

y(1) = e^2

y(1) = c e^(-3) + (2/3) + (1/4)e + (7/9) = e^2

c = e^5 - (1/4)e^4 - (13/9)e^3

confidence rating #$&*:

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

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Given Solution:

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Self-critique (if necessary):

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Self-critique rating:

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Question:

6. The general solution to the equation y ' + p(t) y = g(t) is y = C e^(-t^2) + 1, t > 0. What are the functions p(t) and g(t)?

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Your solution:

y = C e^(-t^2) + 1

ye^(t^2) = C + e^(t^2)

(ye^(t^2))' = 2te^(t^2)

Power rule

y 'e^(t^2) + 2yte^(t^2) = 2te^(t^2)

Remove u(t) which was e^(t^2)

y ' + 2t y = 2t

p(t) = 2t

g(t) = 2t

Confidence rating: 2

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Given Solution:

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Self-critique (if necessary):

Self-critique rating:

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Self-critique (if necessary):

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Self-critique rating:

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Question:

6. The general solution to the equation y ' + p(t) y = g(t) is y = C e^(-t^2) + 1, t > 0. What are the functions p(t) and g(t)?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

y = C e^(-t^2) + 1

ye^(t^2) = C + e^(t^2)

(ye^(t^2))' = 2te^(t^2)

Power rule

y 'e^(t^2) + 2yte^(t^2) = 2te^(t^2)

Remove u(t) which was e^(t^2)

y ' + 2t y = 2t

p(t) = 2t

g(t) = 2t

Confidence rating: 2

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Given Solution:

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Self-critique (if necessary):

Self-critique rating:

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Self-critique (if necessary):

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Self-critique rating:

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