Question

second-order linear ordinary differential equation

y''-y = -2e^(-t)

particulair solution

yp = A*e^(-t)
yp' = -A*e^(-t)
yp'' = A*e^(-t)

If i substitute it in the equation y''-y
I get:

A*e^(-t) - A*e^(-t) = -2*e^(-t)
but the left side is zero?

Answer 1

\[y_P=Ate^{-t}\]

Answer 2

yeah but why?

Answer 3

wolfram says it.. but how can you calculate it

Answer 4

\[\mbox{because term }e^{-t}\mbox{ already exist in homogenic solution: }y_H=C_1e^t+C_2e^{-t}\]

Answer 5

\[r^2-1=0 => r=\pm 1\]

Answer 6

yeah i already got that one :)

Answer 7

y = c1e^x + c2e^-x

Answer 8

but the particulair one... drives me crazy

Answer 9

it should be multiplied by t to make the particular solution linearly independent

Answer 10

ok cool did you try
\[y_p=Ae^{-t}+Bte^{-t}\]
\[y_p'=-Ae^{-t}+Be^{-t}-Bte^{-t}\]
\[y''_p=Ae^{-t}-Be^{-t}+Bte^{-t}-Be^{-t}\]

so
\[y''-y=e^{-t}(A-2B)+Bte^{-t}-[Ae^{-t}+Bte^{-t}]\]
\[=e^{-t}(A-2B-A)+te^{-t}(B-B)=e^{-t}(-2B)\]
this side is suppose to be equal to -2e^{-t}
so -2B=-2 => B=1

Answer 11

but ... then

yp = (At+b)*e^(-t)?
or

yp = t*e^(-t)
and do i have to take the derivative of this to substitute it

Answer 12

and we didn't even need to plug in that homogeneous part i don't know why i did because that parts gonna turn out 0

Answer 13

the second one jessica

Answer 14

because remember be^{-t} is part of your homogenous solution

Answer 15

its gonna end up zeroing out

Answer 16

do you know what i mean?

Answer 17

okee... so yp will be (at+b)e^(-t)

Answer 18

just like what you did :) but a and b are vice versa

Answer 19

you can do that jessica but what i'm saying is you only need ate^{-t}
since be^{-t} is part of your homogenous solution

Answer 20

ahhh oke.... thank you all for you explenations :)

Answer 21

have a good day jessica
i think i will be leaving soon

Answer 22

yeaahh thank you :) i really appreciate your explanations :) i also wish you a good day :)