Question

Can anyone help with integrating factors!!!!

Answer 1

whts the problem

Answer 2

i'm not sure how i would do

dy/dt− 2y = 3e^(2t)

Answer 3

integrating factor is \[e^{\int{-2}dx}\]

Answer 4

is it dt or dy, i'm confused cause i'm not working with a variable x

Answer 5

\[=e^{-2x}\]

Answer 6

im sorry x is t

Answer 7

i'm just not sure how to work out the equation inclusing the integrating factor ..

Answer 8

so the integrating factor is \[e^{-2t}\]

Answer 9

ok i understand that much, so i just put that in to both sides?

Answer 10

\[{d\over dt}(y.e^{-2t})=3e^{2t}.e^{-2t}\]

Answer 11

\[ye^{-2t}=\int{3}dt\]

Answer 12

\[ye^{-2t}=3t+c\]
\[y={e^{2t}(3t+c)\]

Answer 13

\[y=e^{2t}(3t+c)\]

Answer 14

thank you!

Answer 15

Integrating factor equations aren't too difficult once you get used to them

Answer 16

i know im just having a little bit of a problem! thanks yout hoguh!

Answer 17

that is alright, keep em coming,

Answer 18

can i ask for help with the other one that you helped me with ..

Answer 19

dy/dt=−2y+2t−1

for that one, i know the integrating factor .. but am getting lost with putting it in, i kind of understand the one above, but not so much this one.

Answer 20

ok
\[dy/dt=−2y+2t−1\]
make it in the right form
\[{dy\over dt}+2y=2t-1\]
integrating factor is \[e^{\int{2}dt}=e^{2t}\]

\[{d\over dt}{y.e^{2t}=(2t-1)e^{2t}\]

Answer 21

i got that far! its just the next step confuses me!

Answer 22

\[{d\over dt}(y.e^{2t})=(2t-1)e^{2t}\]

Answer 23

ok the next step is to integrate with respect to t
\[\int {d\over dt}(y.e^{2t})dt=\int (2t-1)e^{2t}dt\]

Answer 24

\[ye^{2t}=\int(2t-1)e^{2t}dt\]

Answer 25

oh ok! that is how you do it! ok, i was doing it kinda right!

Answer 26

NB: do not for get the constant of integration

Answer 27

ok i feel stupid, but your really good at this stuf ... will you hate me if i ask another question?

Answer 28

i'm having a hard time finding equilibrium points for a DE with x and y variables

Answer 29

dx/dt= 2x(1 − x/3) − xy
dy/dt= −2y + 3xy

Answer 30

um , i dont really know what you mean by equilibrium points

Answer 31

uhm not points, just equilibriums for each

Answer 32

its when
dx/dt= f(x,y)
dy/dt= g(x,y)

f(x,y)=0
g(x,y)=0

Answer 33

.. if that makes any sense

Answer 34

ok
\[dx/dt= 2x(1 − x/3) − xy=0\]
\[0=2x-{2\over3}x^2-xy\]

Answer 35

but i think its when the general solution is equal to 0 .. not the DE

Answer 36

idk, i hate this :(

Answer 37

i dont feel confident with the question and i dont want to take you in the wrong direction.

i would be happy to help with any other integrating factor questions

Answer 38

would you be able to help me figure out the general solutitons to both
dx/dt= 2x(1 − x/3) − xy
dy/dt= −2y + 3xy
for the first one i want to find x equals ... and just act as if y is a constant,
and then for the second, what y equals, and act as if x is a constant