Question

x''+x' = 100sin5t
Find a particular solution Xp(t) by using the complex function method

Answer 1

100 sin 5t = Im 100 e^{i5t}. So guess xp=Ae^{5it} plug it into the DEQ, solve for A,
and then take Im xp.

Answer 2

@fwizbang dude more inf please

Answer 3

Can I reply :) .?!

Answer 4

@Neemo of course :)

Answer 5

Ok :) ! "A" here is a complex number ! \[xp'(t)=5iAe^{i5t}\]
then \[xp"(t)=-25Ae^{i5t}\]
so we should have \[(5Ai-25A)e^{i5t}=100e^{i5t}\]

Answer 6

did you understand, ?should I continue ? @SkykhanFalcon

Answer 7

@Neemo ok i understand

Answer 8

then A= ???!

Answer 9

@Neemo wait how about 5Ai ?

Answer 10

sorry ! :) xp'(t)=5Aie^...did you undertand this ?

Answer 11

@Neemo yeah i understand that i mean that (5Ai-25A) = 100 how can we know A

Answer 12

\[5A(i-5)=100\]
then \[A(i-5)=20\]

Answer 13

oh well A is complex

Answer 14

you are right

Answer 15

dude something wrong i guess

Answer 16

why ?!

Answer 17

@neemo the answer should be (-10/13)*(5sin5t+cos5t)

Answer 18

the book says so

Answer 19

we haven't finished yet ! :) !

Answer 20

so can you write A=x+iy ?

Answer 21

of course i can @Neemo

Answer 22

-(20i+100)/26

Answer 23

It seems correct ! now can you compute Im(xp(t))=Im(Ae^i5t)

Answer 24

@neemo how can i do that :/ i will be here in 20 minutes please :) something came up please wait :)

Answer 25

kk I'll be here !don't worry !

Answer 26

@Neemo ok here i am..you mean -20/26 = Ae^î5t ?

Answer 27

back ! @SkykhanFalcon

Answer 28

no ! \[xp(t)=Ae^{5 i t}=A(\cos(5t)+isin(5t))\]
then compute Im(xp(t)) with A =-(20i+100)/26=-(10/13)i-(50/13)

Answer 29

stiill there @SkykhanFalcon ?!

Answer 30

ok i understand :) thank u so much :)

Answer 31

Your're welcome anytime ! When you have an equation like x"+..x'+x=..cos(??t)
then choose the Re(xp(t))

Answer 32

ok i get it thank u again :)