query_02

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course MTH 279

******THIS ENTIRE QUERY FOOLED ME....NOT SURE IF I SHOULD HAVE SOLVED THESE EQUATIONS AS SEPARABLE EQUATIONS?!********Solve each equation:

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Question: 1. y ' + y = 3

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Your solution:

y' + y = 3

This equation is nonhomogeneous, as it does not equal 0, but instead equals 3 and is in the form of y' + p(t)y = g(t).

Here we can tell that p(t) = 1 and that g(t) = 3. An antiderivative is P(t) = t

We let P(t) be some antiderivative of p(t), and define a new function u(t) by u(t) = e^(P(t)), or u(t) = e^t

Multiplying the diff. eq. by u(t), we get (e^t)y' + (e^t)y = 3(e^t) + C

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e^t is your integrating factor.

(e^t)y' + (e^t)y is the derivative with respect to t of y e^t.

So we can integrate both sides with respect to t.

By the first sentence the integral of the left-hand side is y e^t + c.

You should be able to complete the solution by integrating the right-hand side to get an equation you can solve for y.

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confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary): Where do I go after this? Should I have solved this as a separable equation??

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It is possible to solve this as a separable equation, but since this assignment is attached to a section that doesn't mention separable equations, you shouldn't.

Many of the equations you see in this assignment can be solved as separable, but that's not the technique being developed here.

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Self-critique rating:

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Question:

2. y ' + t y = 3 t

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Your solution:

p(t) = t

g(t) = 3t

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

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

y'(e^((t^2)/2)) + ty(e^((t^2)/2)) = 3t(e^((t^2)/2))

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The left-hand side is, as before, the derivative of y multiplied by a function of t. So the left-hand side can be integrated with respect to t.

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confidence rating #$&*:

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Given Solution:

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

Self-critique (if necessary): Should I have solved this as a separable equation?!

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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:

p(t) = -4

g(t) = sin(2t)

P(t) = -4t

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

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

confidence rating #$&*:

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Given Solution:

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

Self-critique (if necessary):Should I have solved this as a separable equation?!

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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:

p(t) = 1

P(t) = t

u(t) = e^t

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

confidence rating #$&*:

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Given Solution:

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

Self-critique (if necessary): Should I have solved this as a separable equation?!

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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:

p(t) = 3

P(t) = 3t

u(t) = e^3t

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

confidence rating #$&*:

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Given Solution:

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

Self-critique (if necessary): Should I have solved this as a separable equation?!

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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:

To find the equation (which will give us p(t) and g(t)), we will work backwards.

It is obvious we divided by e^(t^2), so we know that after we multiplied the equation by u(t), the equation was (e^(t^2)) y = 1(e^(t^2)) + C.

The equation before it was simplified could have been (e^(t^2))y' + te^(t^2)y = 2te^(t^2). Therefore we know that u(t) = e^t^2.

Since u(t) = e^P(t), we know that P(t) = t^2. We know that P(t) is an antiderivative of p(t), so p(t) = 2t and g(t) = t.

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary): Should I have solved this as a separable equation?!

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Self-critique rating:"

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You are correctly finding your integrating factor and applying it to the equation.

There's one step you're missing.

Check my notes then see if you can complete your solutions.

&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).

Be sure to include the entire document, including my notes.

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