Question

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)?

Answer 1

http://www.sosmath.com/tables/diffeq/diffeq.html

Answer 2

yes you are right so far

Answer 3

k thanks, i messed up the math somewhere then ill have to redo it.

Answer 4

that site is nice above

Answer 5

ok you know we only found the homogeneous solution right?

Answer 6

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?

Answer 7

i assume you accidently left out the cos of that constant thingy
i will check your solution in just a sec k?

Answer 8

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

Answer 9

the nonhomogeneous is y=sin(2t)

Answer 10

wait....lol

Answer 11

i think thats right
its just weird to see the nonhomogeneous solution as a part of the homogenous solution

Answer 12

let me see what i come up with doing what you which i think you tried y=Atcos(2t)
right?

Answer 13

i think i made a mistake in checking it because your solution is what i just found sorry

Answer 14

did you try checking it?

Answer 15

haha yes. i think its all set i went back over the work. man i miss diff eq. that class was interesting.

Answer 16

lol

Answer 17

Refer to the attachment.

Answer 18

i'm confused why are you doing t's and x's?

Answer 19

Well is my face red. Mortified.
The second solution version is attached.
Everything is in terms of the independent variable t now.

Answer 20

Some of the more complex expressions could be further simplified.