anonymous
  • anonymous
y''+4y=2sin(2t) y(0)=1 y'(0)=-1solve the I.V.P. the complex root threw me. is it c1cos(2t) +c2sin(2t)?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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myininaya
  • myininaya
http://www.sosmath.com/tables/diffeq/diffeq.html
myininaya
  • myininaya
yes you are right so far
anonymous
  • anonymous
k thanks, i messed up the math somewhere then ill have to redo it.

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myininaya
  • myininaya
that site is nice above
myininaya
  • myininaya
ok you know we only found the homogeneous solution right?
anonymous
  • anonymous
yeah i can do the rest i was just unsure about the complex root and setting up the homogeneous solution. i get y(t)=c1(2t) +c2sin(2t)-(1/2)tcos(2t) for the gen solution. that seem right?
myininaya
  • myininaya
i assume you accidently left out the cos of that constant thingy i will check your solution in just a sec k?
myininaya
  • myininaya
ok your solution is not checking out i tried it twice maybe a maybe a mistake but let me try actually getting the nonhomogeneous k? thats the only thats not working out right now
myininaya
  • myininaya
the nonhomogeneous is y=sin(2t)
myininaya
  • myininaya
wait....lol
myininaya
  • myininaya
i think thats right its just weird to see the nonhomogeneous solution as a part of the homogenous solution
myininaya
  • myininaya
let me see what i come up with doing what you which i think you tried y=Atcos(2t) right?
myininaya
  • myininaya
i think i made a mistake in checking it because your solution is what i just found sorry
myininaya
  • myininaya
did you try checking it?
anonymous
  • anonymous
haha yes. i think its all set i went back over the work. man i miss diff eq. that class was interesting.
myininaya
  • myininaya
lol
anonymous
  • anonymous
Refer to the attachment.
myininaya
  • myininaya
i'm confused why are you doing t's and x's?
anonymous
  • anonymous
Well is my face red. Mortified. The second solution version is attached. Everything is in terms of the independent variable t now.
anonymous
  • anonymous
Some of the more complex expressions could be further simplified.

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