Query 02

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course Mth 279

6/16 12

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

Question: 1. y ' + y = 3

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Your solution:

y = e^-P(t) * int[ e^P(t) * g(t) ]dt + C*e^-P(t)

g(t) = 3

p(t) = 1

P(t) = t

y = e^-t * int(e^t * 3)dt + C*e^-t

y = e^-t * 3e^t + C*e^-t

y = C*e^-t + 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:

2. y ' + t y = 3 t

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Your solution:

y = e^-P(t) * int[ e^P(t) * g(t) ]dt + C*e^-P(t)

g(t) = 3t

p(t) = t

P(t) = t^2/2

y = e^(-t^2/2) * int(e^(t^2/2) * 3t)dt + C*e^(-t^2/2)

y = e^(-t^2/2) * int(3te^(t^2/2))dt + C*e^(-t^2/2)

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

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

@&

You appear to be using a formula that you're unlikely to remember, as opposed to the integrating factor technique.

I recommend the technique over a quickly-forgotten formula.

*@

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:

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

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Your solution:

y = e^-P(t) * int[ e^P(t) * g(t) ]dt + C*e^-P(t)

g(t) = sin(2t)

p(t) = -4

P(t) = -4t

y = e^-(-4t) * int(e^(-4t) * sin(2t))dt + C*e^-(-4t)

y = e^4t * int(e^(-4t) * sin(2t))dt + C*e^4t

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

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

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

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:

y = e^-P(t) * int[ e^P(t) * g(t) ]dt + C*e^-P(t)

g(t) = e^t

p(t) = 1

P(t) = t

y = e^-t * int(e^(t) * e^t)dt + C*e^-t

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

y = C*e^-t + e^t/2

inital condition

2 = C*e^0 + e^0/2

2 = C +1/2

C = 3/2

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

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:

y = e^-P(t) * int[ e^P(t) * g(t) ]dt + C*e^-P(t)

g(t) = e^t + 2t + 3

p(t) = 3

P(t) = 3t

y = e^-3t * int(e^3t * (e^t + 2t + 3))dt + C*e^-3t

y = e^-3t * int( (e^t)^4 + 2t(e^t)^3 + 3(e^t)^3 )dt + C*e^-3t

y = e^-3t * [((e^t)^4)/4 + (2t(e^t)^3)/3 + (7(e^t)^3)/9 ] + C*e^-3t

y = e^t/4 +2t/3 + 7/9 + C*e^-3t

y = C*e^-3t + 2t/3 + e^t/4 + 7/9

initial condition

e^2 = C*e^(-3(1)) + (2(1))/3 + e^1/4 + 7/9

e^2 = C*e^-3 + 2/3 + 1/4 +7/9

C = e^5 - (61e^3)/36

y = (e^5 - (61e^3)/36)*e^-3t + 2t/3 + e^t/4 + 7/9

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:

Looking at the solution and comparing to the standard format of C*e^-P(t)

P(t) = t^2

p(t) = 2t

In order to have a the constant 1 in the solution the e^-P(t) term must be multiplied by the integral of e^P(t)*g(t)

and equal one.

We know e^-P(t) = e^-t^2 and e^P(t) = e^t^2

Therefore 1 = e^-t^2 * int (e^t^2 * g(t))

int(e^t^2 *g(t))dt = e^t^2

e^t^2 * g(t) = d/dt(e^t^2)

e^t^2 * g(t) = 2t*e^t^2

g(t) = 2t

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

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Self-critique rating:"

@&

Good, but you're relying on a formula rather than a technique.

You should use the technique.

Check my note.

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