Question

help!!!

solve the differential eqn: x''-4x=e^(-2t)

given x(0)=0 and x'(0)=1

Answer 1

what's the complimentary?

Answer 2

compl= Ae^(2t)+Be^(-2t)

Answer 3

so what is your guess for the particular?

Answer 4

i tried x= Ae^(-2t)
x'=-2Ae^(-2t)
x''=4Ae^(-2t)

Answer 5

you can't do that because we already have a term Ae^(-2t) in the complimentary, and we can't have any parts of the complimentary and the particular linearly dependent

Answer 6

how do we normally deal with two linearly dependent solutions, like in a double root?

Answer 7

ow... that's why... there's a rule? wow

Answer 8

we add a "t" to C2

Answer 9

yes, the rule is that no terms y1, y2,... ,yn or yp can be linearly dependent
if they are linearly dependent we have to multiply something by the independent variable usually

Answer 10

add a t to the guess for the particular in this case

Answer 11

x= Ate^(-2t)
x'=-2Ate^(-2t)
x''=4Ate^(-2t)

Answer 12

careful, you need to use the product rule in that derivative

Answer 13

ow yes... wait

Answer 14

x= Ate^(-2t)
x'=Ae^(-2t)-2tAe^(-2t)
x''=-2Ae^(-2t)-2Ae^(-2t)+4Ate^(-2)

Answer 15

yes

Answer 16

:) i'm getting better at this

A=-1/4??

Yp=(-1/4)te^(-2t)

Answer 17

yes you are, that's the right particular :)

Answer 18

now my initial conditions

x(0)=0 gives A+B=0

Answer 19

yes, and the other condition?

Answer 20

A=-5/16
B=5/16

Answer 21

yes, nice!

Answer 22

actually I think you switched A and B, but besides that, nice job

Answer 23

LOL... i just realized that... i couldn't find my x(1)... in one of the MCQs given... but i got it now.. thanks

Answer 24

welcome!

Answer 25

i actually learnt a lot from u since last night

bernouli eqn
particular solns... for a linear i use mx+c, quadratic ax^2+bx+c, one that has the same as my particular soln i add a t :)

Answer 26

one last question before i go to bed :)