Question

help me in linear diff equation with only variables given

Answer 1

problem is \[\frac{ dv }{ dt }+\frac{ k }{ m} v = \pm g\]
g is constant;

Answer 2

@myininaya

Answer 3

@ganeshie8 @welshfella

Answer 4

Hint :
It is separable

Answer 5

oh so multiply both side by dt and m?

Answer 6

am i right?

Answer 7

Give it a try and see what you get

Answer 8

i got \[\frac{ dv }{ m } + \frac{ k }{ dt } = \frac{ \pm g }{ (dt)(m)}\] @ganeshie8

Answer 9

What good is that ?

Answer 10

You didn't multiply. o.O

Answer 11

Maybe just go with the linear thingy
try finding IF

Answer 12

Because the separable method involves painful algebra here

Answer 13

\[[\frac{ dv }{ dt } + \frac{ k }{ m } v = \pm g] (dt)(m)\]

Answer 14

problem is \[\frac{ dv }{ dt }+\frac{ k }{ m} v = \pm g\]
g is constant;

Answer 15

The equation is already in standard form. See if you can find the IF

Answer 16

so p is k/dt and q is g

Answer 17

P(t) is simply k/m here

Answer 18

oops yes \[e ^{\int\limits \frac{ k }{ m }}\]

Answer 19

First, convince yourself that the equation is in variables v and t
not x and y

Answer 20

again, integral sign must always end with a differential

Answer 21

The expression \[e ^{\int\limits \frac{ k }{ m }}\]

makes no sense

Answer 22

dm

Answer 23

\[e ^{\int\limits \frac{ k }{ m }} dm\]

Answer 24

\[\LARGE IF = e ^{\int\limits \frac{ k }{ m }\,\color{red}{dt}}\]

Answer 25

k, m , g are constants here

Answer 26

ahsorry. cause were dt and dv

Answer 27

v and t are the variables

Answer 28

yes

Answer 29

yes

Answer 30

i dont know how to integrate only variables O.O

Answer 31

k,m are constants. Just pull them out of the integral

Answer 32

\(\frac{k}{m}t\)?

Answer 33

so kt/mt

Answer 34

\[\LARGE IF = e ^{\int\limits \frac{ k }{ m }\,\color{red}{dt}} = e ^{\frac{ k }{ m }\int\limits 1\,\color{red}{dt}} \]

Answer 35

okay so our if is just t

Answer 36

Yes

\[\LARGE IF = e ^{\int\limits \frac{ k }{ m }\,\color{red}{dt}} = e ^{\frac{ k }{ m }\int\limits 1\,\color{red}{dt}} = e^{(\frac{k}{m})t}\]

Answer 37

so IF = t now?

Answer 38

:/

Answer 39

why

Answer 40

Stare at this

\[\LARGE IF = e ^{\int\limits \frac{ k }{ m }\,\color{red}{dt}} = e ^{\frac{ k }{ m }\int\limits 1\,\color{red}{dt}} = e^{(\frac{k}{m})t}\]

Answer 41

\(IF = \Large e^{\frac{k}{m}t}\)

Answer 42

stare at that for few minutes and tell me why you think the IF is t

Answer 43

This time it's not \(e^{\ln}\)

Answer 44

ah yes, so IF is just e^k/m (t)

Answer 45

\[(e ^{\frac{ k }{ m }t} ) v = \pm g(e ^{\frac{ k }{ m }t})\]

Answer 46

@ganeshie8

Answer 47

\(\Large(e^{\frac{k}{m}t})v = \int \pm g(e^{\frac{k}{m}t})dt+c\)

Answer 48

@Sachintha can u finished it

Answer 49

@johnweldon1993

Answer 50

@inkyvoyd

Answer 51

@sleepyjess

Answer 52

@Qwertty123

Answer 53

Do you know to keep working on this?

Answer 54

nope

Answer 55

okay, Not sure either... @ShadowLegendX @inkyvoyd

Answer 56

\[\frac{ dv }{ dt } + \frac{ k }{ m } v = \pm g\]
\[P(t) = \frac{ k }{ m }\]
\[Q(t) = \pm g\]
IF. \[e ^{\int\limits \frac{ k }{ m }} dt\]
\[e ^{\int\limits \frac{ k }{ m }t} \]

Answer 57

whats next

Answer 58

\(\large\begin{array}{ccc}&\int \pm g(e^{\frac{k}{m}t})dt&=&\pm g\int (e^{\frac{k}{m}t})dt\\&&=&\pm g(e^{\frac{k}{m}t})(\frac{k}{m})+c\end{array}\)

Not sure whether I have done it correctly.

Answer 59

but the answer in the book should be \[v = ce ^{\frac{ -km }{ t }} \pm \frac{ mg }{ k }\]

Answer 60

@inkyvoyd help pls

Answer 61

Well if you rearrange the earlier equation with the result of the integration, it might get close to the final answer.

Answer 62

pls help.....................

Answer 63

\(\begin{array}{ccc}&\large(e^{\frac{k}{m}t})v&=&\large\int \pm g(e^{\frac{k}{m}t})dt+c\\&&=&\large\pm g(e^{\frac{k}{m}t})(\frac{k}{m})+c\\&\Large\frac{(e^{\frac{k}{m}t})v}{(e^{\frac{k}{m}t})}&=&\Large\frac{(\pm g(e^{\frac{k}{m}t})(\frac{k}{m})+c)}{(e^{\frac{k}{m}t})}\\&\large v&=&\large ce^{-\frac{k}{m}t}\pm g\frac{k}{m}\end{array}\)

Answer 64

Are you sure that you have written the answer of the book correctly?

Answer 65

yes i do right ciorrectly

Answer 66

I didn't find any errors with my calculation. Might have overseen it. @SnuggieLad please check what's wrong. :)

Answer 67

@Sachintha I know I'm a little late to the party, but real quick

\[\large \int e^{\frac{k}{m}t}dt = ?\]

\(\large u = \frac{k}{m}t\) so \(\large du = \frac{k}{m}dt\) meaning \(\large dt = \frac{m}{k}du\)

Everything else looks great!

Answer 68

Oops. So that's where I have gone wrong. I seemed to have differentiated it instead of integrating. -_- Pardon me.

Answer 69

\(\large\begin{array}{ccc}&\int \pm g(e^{\frac{k}{m}t})dt&=&\pm g\int (e^{\frac{k}{m}t})dt\\&&=&\pm g(e^{\frac{k}{m}t})(\frac{m}{k})+c\end{array}\)

Answer 70

\(\begin{array}{ccc}&\large(e^{\frac{k}{m}t})v&=&\large\int \pm g(e^{\frac{k}{m}t})dt+c\\&&=&\large\pm g(e^{\frac{k}{m}t})(\frac{m}{k})+c\\&\Large\frac{(e^{\frac{k}{m}t})v}{(e^{\frac{k}{m}t})}&=&\Large\frac{(\pm g(e^{\frac{k}{m}t})(\frac{m}{k})+c)}{(e^{\frac{k}{m}t})}\\&\large v&=&\large ce^{-\frac{k}{m}t}\pm g\frac{m}{k}\end{array}\)

Answer 71

There we go! :)

Answer 72

Thanks for showing the mistake I have done. :)