General Solution of the first order differential equation

  • Thread starter Yr11Kid
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  • #1
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Main Question or Discussion Point

dy/dt + y = [tex]\infty[/tex] [tex]\sumSin(nt)/n^2[/tex] n=1


Ok still a bit new with all these symbols and stuff but that is the basic jist of it.

y(t) = yh(t) + yp(t) it what i thought about using to start off with, yh(t) = Acos2t + Bsin2t.
Then subbing yp(t) into the differential equation. Not really sure about how to do this, help much appreciated :)
 

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  • #2
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Hey everyone,
This question kind of confused me, i think the general idea is to use y(t) = yh(t) + yp(t) and basically saying yh(t) = Acos2t + Bsin2t

to find yp(t) subbing into the differential equation and equating.

help much appreciated :)
 

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  • #3
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Here's an idea. It shouldn't be too hard to find particular solutions to dy/dt + y = sin(nt)/n^2, where n is an unspecified constant. Do that and then sum the solutions. You'll end up with a weird infinite series, but it should be convergent because of the n^2 in the denominator. And this is probably the best you can do, since the problem itself has a series as the nonhomogeneous term.
 
  • #4
hunt_mat
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Do you mean:
[tex]
\frac{dy}{dt}+y=\sum_{n=1}^{\infty}\frac{\sin (nt)}{n^{2}}
[/tex]
This is an integrating factor question.
 
  • #5
HallsofIvy
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The general solution to the equation y'+ y= 0 is [itex]y(t)= Ce^{-t}[/itex]

To find the general solution to the entire equation, look for a solution of the form
[tex]\sum_{n=1}^\infty A_n sin(nt)+ B_n cos(nt)[/tex]

That will give you as sequence of equations to solve for [itex]A_n[/itex] and [itex]B_n[/itex]. Once you have that, add to [itex]Ce^{-t}[/itex].
 
  • #6
HallsofIvy
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I am combining the two threads on the same thing.
 

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