Query2

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course MTH 279

2/19/12 6:25

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

Question: 1. y ' + y = 3

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Your solution:

We know that p(t) = 1 , so P(t) = -int 1 dt => t

So u(t) = e^(-t) and u’(t) = -e^(-t) u(t) = P(t)

To get our general solution we multiply u(t) by both sides.

u(t)y’ + u(t)p(t)y = u(t) * 3 p(t) = 1 and u(t)p(t) = u’(t) which gives us the equation

u(t)y’ + u’(t)y = u(t) * 3

[u(t)y(t)]’ = u(t) * 3 now we integrate both sides and obtain

u(t)y(t) = integral u(t)*3

e^(-t)y(t) = integral 3e^(-t)

e^(-t)y(t) = -3e^(-t) + C set = to y(t) and we obtain;

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

it think is what you wanted.

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P(t) = integral(1 dt) = t, so u(t) = e^t.

If you were solving y ' + y = 0 you would get y = A e^-t.

The solution to this problem would be

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

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Given Solution:

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Question:

2. y ' + t y = 3 t

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Your solution:

p(t) = t P(t) = t^2/2

Integrating factor = e^(t^2/2) = u(t)

u(t) y’ + u(t) p(t) y = u(t) 3t

(u(t) y)’ = u(t) 3t

(e^(t^2/2) y)’ = e^(t^2/2)3t Int/ both sides

e^(t^2/2)y = 3 e^(t^2/2) + c set = to y

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

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Given Solution:

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Question:

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

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Your solution:

p(t) = -4 P(t) = -4t

Integrating factor = e^-4t = u(t)

e^-4t * y’ + e^-4t * (p(t)) * y = e^-4t(Sin(2t))

(e^-4t y)’ = e^-4t(sin(2t)) int both sides

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

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

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Good, but use the order of operations correctly (in this case group your numerator). If you type that into a computer algebra system you won't get what you're expecting.

(-cos(2t) + 2sin(2t) ) /10 + c e^4t

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Given Solution:

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Question:

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

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Your solution:

p(t) = 1 P(t) = t integrating factor = e^t = u(t)

(e^t y)’ = e^t * e^t

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

y= e^t/2 + ce^(-t)

y(0) = 2

½ + C = 2 C = 3/2

So our equation is

Y = e^t/2 + 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:

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

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

y(1) = (9e + 52)/36 + ce^(-3) = e^2

= 2.12401 +c *.049787 = 7.38906

C* .049787 = 5.26505

C = 105.752

Y = (9e^t + 52t)/36 + 105.752e^(-3t)

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Given Solution:

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Self-critique (if necessary):

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c = (e^2 - (9 e + 52) / 36) * e^3

which does approximate to 105.752.

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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:

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Given Solution:

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Self-critique (if necessary):

I’m not sure how I’m supposed to do this… I deduced that p(t) could be 2t… but I’m not sure how to find g(t).

I believe you showed us how to do this in class, and I remember taking notes on it.. but for some reason, I can’t locate them…

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Self-critique rating:

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The integrating factor will be e^integral(p(t) dt), so your equation will after a step or two become

e^(integral(p(t) dt) ) * y = integral(g(t) dt) + C

giving you

y = C e^(-integral(pIt) dt) + integral(g(t) dt) / e^(integral(pIt) dt)

The first term on the right-hand side is said to be C e^(-t^2), which you can use to figure out p(t).

The second term is 1, so p and g must be the equal functions.

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Not bad. Check my notes and be sure you understand everything. Let me know if not.

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