Query 02

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course Mth 279

reminder to send some link that you said you would send last time but had a cpu malfunction.2/26 8

query 2

Solve each equation:

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Question: 1. y ' + y = 3

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Your solution:

separate the sides dy(t)/dt = 3-y(t), now divide both sides by (3-y(t)) to get into an easy form to integrate.

your left with ( dy(t)/dt/3-y(t) ) dt = 1 dt

integrate to get log(3-y(t)) = t + c

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

solve for y(t) to get,

y(t) = Ae^-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:

get each side to an appropriate integration solution

dy(t) / dt = 3t - ty(t) ; simplify by taking t out on right side

dy(t) / dt = t(3 - y(t) ) ; now divide by (3-y(t)) on both sides for integration

[ dy(t) / dt / (3-y(t)) ] dt = t dt

log (3-y(t)) = t^2/2 + c; simplify more

3-y(t) = e^(t^2/2 + c); now solve for y(t) to get

y(t) = Ae^-(t^2/2) + 3

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Your solutoin by separating variables is pretty good, but your final function y(t), when substituted into the equation, gives you

0 = 3 t.

Check that yourself, because I can't immediately spot an error in your steps (except that you aren't treating this as a linear equation, which you should be doing on this assignment).

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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:

working from example 1 basically in Chapter 2.3, I have widdled this one down a bit.

let u = e^-4t and multiply each side.

then by substituting d/dt(e^-4t) = -4y(t)e^-4t

now this is the reverse product rule from which i had to look up. Now we can integrate this, in theory;

int ( d/dt e^-4t y(t) )dt = int ( e^-4t sin2t ) dt

?? I had a lot of difficulty with this integration, later I found that it is actually a rule; which states;

int exp(alpha t) sin ( beta t) dt = exp(alpha t)(-beta*cos(beta t)) + alpha*sin(beta t)) / alpha^2 + beta^2

I still could not sipher beta and alpha with e^-4t value. But my answer ended up to be

y(t) = -1/10 (2sin(2t) + cos(2t))

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This is obtained by straightforward integration by parts. It's in a table but you need to be able to apply integration by parts (demonstrated in class today on a similar question).

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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:

let u = e^t because p(t) = 1. and P(t) = t, so u will be e^t

now multiply each by u to get;

e^t dy(t)/dt + e^t y(t) = e^2t

now apply the inverse product rule to the left side which states f(dg/dt) + g(df/dt) = d/dt(fg)

this simplifies to d/dt e^t y(t) dt = e^2t dt

the integration becomes e^t*y(t) = e^(t^2/2) + c

simplified to solve for y(t) = e^(t^2/2) + ce^-t

so with our initial condition y(0) = 2, our equation

e^0 + ce^-(0) = 2

c = 3/2

now fill into our y(t) equation

y(t) = e^t^2 / 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:

quite involved, going to brush over some typing

p(t) = 3, so P(t) = 3t, which makes our u sub = e^3t

multiply each by u, then you are left with the inverse product rule here to integrate more easily.

when integrating, you get e^3t*y(t) = 2/3 t e^3t + 7/9 e^3t + 1/4 e^4t + c

now solve for y(t) to after getting a common denominator to get

y(t) = 24t + 28 + 9e^t + c*e^-3t

solve for c to get that c = -52 - 8e^2 / e^3

fill back into the origional equation y(t) to get that y(t) = 24t + 28 + 9e^t + (-52 - 8e^2 / e^3)e^-3t

i'm sure it can be simplified some.

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You followed the right method. You did leave out some steps, but the solution is OK. Try to fill in the steps, though, so you'll really understand what you're doing.

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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:

p(t) = -2t

and g(t) = 0

I know this because looking at every other problem i've done, and each example, I always end up with my u sub in there, when my u sub is e^something,

that something started out at an antiderivative of that before it was raised by e.

g(t) = 0 therefore G(t) = 1

p(t) = -2t P(t) = -t^2 u = e^-t^2

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

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Not bad. You did use the right method on some of these, and could easily enough have done so on others.

See my notes and be sure you work through all the assigned problems using the right method. Once you get a few of these, they don't take long

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