Query 02

#$&*

course Mth 279

1/3/14 around 5 pm

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

@&

Good solution using a change of variable.

However the section was on nonhomogeneous linear equations, and you do need to practice those techniques at this point.

Be sure you can solve the equation using an integrating factor.

*@

 

 

confidence rating #$&*:8232; 

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Given Solution: 

 

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Self-critique (if necessary):OK

 

 

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Self-critique rating:3

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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 #$&*:8232; 

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

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

Given Solution: 

 

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Self-critique (if necessary):OK

 

 

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Self-critique rating:3

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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)

 

 

@&

Good. This is the appropriate technique for this section, though as I said earlier your solution by change of variable was also very good.

*@

confidence rating #$&*:8232; 

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

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

Given Solution: 

 

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

Self-critique (if necessary):OK

 

 

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Self-critique rating:3

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

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 #$&*:8232; 

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

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

Given Solution: 

 

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Self-critique (if necessary):OK

 

 

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Self-critique rating:3

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

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 #$&*:8232; 

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

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

Given Solution: 

 

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Self-critique (if necessary):OK

 

 

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Self-critique rating:3

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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)

I used the Power rule here

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

—>

y ' + 2t y = 2t

p(t) = 2t

g(t) = 2t

 

 

confidence rating #$&*:8232; 

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

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

Given Solution: 

 

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Self-critique (if necessary):OK

 

 

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Self-critique rating:3"

&#Your work looks good. See my notes. Let me know if you have any questions. &#