Question

Solve the initial value problem.

dv/dt + rv = -g, where v(0) = v(o)

Answer 1

linear ode?

Answer 2

yeah.. gravitational acceleration, and velocity

Answer 3

yes it is linear and integrating factor swill be used .

Answer 4

\[\frac{dv}{dt} +rv =-g\]
integrating factor: \(u=e^{\int r(t)dt}\)

Answer 5

hmm..i dont think that is right :/

Answer 6

It's dv/dt

Answer 7

yup :P

Answer 8

saw right now .

Answer 9

I assume it's\[\frac{dv}{dt} +r(t)v=-g\]

Answer 10

It is linear!

Answer 11

I hate it when people delete!! :P

Answer 12

me too :D

Answer 13

why dv/dt=g??

Answer 14

its acceleration .

Answer 15

isnt acceleration d^2v/dt^2??

havent done there in a while lol

Answer 16

these*

Answer 17

no derivative of velocity is acceleration .

Answer 18

me too doing after long time !

Answer 19

yeah!! lol..i forgot

Answer 20

but why does acceleration =g?

Answer 21

its i guess related to gravitaional field

g is gravitational acceleration .

Answer 22

i thought g denotes gravity :/

Answer 23

so everyone... thank you so much for sticking with this, i have had some issues going on here at the house... if it helps here is the solution... i am trying to understand how to get there...

v(t) =(−g/r)+(v(0) +g/r)e^−rt

Answer 24

dv/dt + rv = -g, where v(0) = v(o)

v'(t)+v(t)*r=-g

v'(t)=-g-v(t)r

v'(t)
----- =1
-g-v(t)r

Integrate both side

Answer 25

Integral of a'(t)/a(t)= log(a(t))

Answer 26

but why isnt it a linear ode??

Answer 27

but isnt**

Answer 28

i am unsure how you integrate that... integrating the right side is just v(t) right but what about the left

Answer 29

ok here we go .

Answer 30

why did you set h(x) = to the integral of rdt

Answer 31

to get integrating factor.

Answer 32

okay, i see, cuz it is infront of the v in the original equation.

Answer 33

yes exactly.

Answer 34

i am not seeing how you went from your second step... after you multiplied by the integrating factor, to the third line, where you simplified that left hand side.

Answer 35

ok by product rule
\[\Large \frac{d}{dt}(e^{rt}*v)=e^{rt}*\frac{dv}{dt}+v*\frac{d}{dt}(e^{rt})\]
\[\Large \frac{d}{dt}(e^{rt}*v)=e^{rt}*\frac{dv}{dt}+e^{rt}*rv\] which is the left hand side .

Answer 36

i understand when you do that... seems like you are working backwards though, i would of never thought of that.... but the rest of it I understand good. very clear.

the next part of the problem ask for
b) show that lim v(t) = -g/r, as t goes to pos infinity.....

which equation are we suppose to use for this?

Answer 37

its the one that you got before

v = -g/r + e^-rtC right

Answer 38

the last one
as t goes to inginity exponetial terms gets zero you ar left with only -g/r

Answer 39

oh you would use the equation after you substituted the initial velocity? the very last one in red, i mean we would get the same answer as using the one before putting in the initial velocity, so does it really matter?

Answer 40

yes use the red one .its important .

Answer 41

okay... and the last part of this problem asks

c) integrate v to obtain the height y, assuming an initial height y(0) =y(o)

Answer 42

are we integrating the red equation?

Answer 43

why would you integrate ????

Answer 44

i dont know... that is exactly what the problem tells me to do.

Answer 45

thats it you got the soluion which is in red . and you find that v =-g/r when t tends to infinity . that was your question .

Answer 46

yes, that was part of the question too, and the last part says to integrate v to obtain height y.

Answer 47

ok integrate it.

Answer 48

so if i integrate that first term -g/r, and i am finding y(t) then do will it just be -gt/r?