Integrating Factor

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Homework Statement


y' + (2/t)y = (cost)/(t^2), and the following condition is given: y(pi) = 0


Homework Equations





The Attempt at a Solution


After employing the integrating factor, I find the solution to be:

[itex]y=e^{-2t} \int e^{2t} \frac{\cos(t)}{t^2} dt[/itex].

Evidently, this simplifies all the way to y = (sin t)/(t^2). I am not sure as to how this integral should be solved. Any hints would be much welcomed.
 

Answers and Replies

  • #2
Curious3141
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I think you should show all your steps, starting with exactly how you applied your integrating factor. Clearly, there's an error somewhere.
 
  • #3
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I see what I did wrong.

mu of t, the integrating factor, [itex]\mu (t) = e^{\int \frac{2}{t}}dt = e^{2 \ln t} = e^{2t}[/itex]

Do you see where I went wrong? It should be [itex]\mu (t) = 2t[/itex]
 
  • #4
HallsofIvy
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This is an example of what happens when you memorize formulas (imperfectly) rather than learning basic definitions and how the formulas are derived.

An "integrating factor" is a function, [itex]\mu(t)[/itex] such that multiplying the equation by it converts the left side into a single derivative. Here, that means we must have [itex]\mu y'+ (2\mu/t)y= (\mu y)'[/itex]. Expanding the derivative on the right that becomes [itex]\mu y'+ \mu' y= \mu y'+ (2\mu/t)y[/itex] which reduces to [itex]\mu'= 2\mu/t[/itex], a separable differential equation for [itex]\mu[/itex]. [itex]d\mu/\mu= 2dt/t[/itex] integrates to [itex]ln(\mu)= 2ln(t)[/itex] or [itex]\mu= t^2[/itex] NOT "2t" (I have neglected the "constant of integration since we only need a single function).

Multiplying the entire equation by [itex]t^2[/itex] gives [itex]t^2y'+ 2ty= (t^2y)'= cos(t)[/itex] which is easy to integrate.
 
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  • #5
Curious3141
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I see what I did wrong.

mu of t, the integrating factor, [itex]\mu (t) = e^{\int \frac{2}{t}}dt = e^{2 \ln t} = e^{2t}[/itex]

Do you see where I went wrong? It should be [itex]\mu (t) = 2t[/itex]
##e^{2 \ln t} = (e^{\ln t})^2 = t^2##, that's your integrating factor.

Your integrating factor is NOT ##2t##. ##2t## is in fact the derivative of your integrating factor, and you should be able to see this from applying product rule to ##yt^2##.
 
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