Question

let's say we have a function F(y(t),t)=y*G(t)
the partial derevative of F with respect to t
is y*dG/dt or dy/dt*G+dG/dt*y
I would really apreciate an explanation

Answer 1

option 2

Answer 2

since both are function of t!

Answer 3

\[F(y(t), t) = y \times G(t)\] ?

Answer 4

yes

Answer 5

\[\frac{ ∂G }{ ∂t }\]

Answer 6

yes

Answer 7

if both G and Y are function of t.. then you have to use the product rule!

Answer 8

mashy our teacher used the first one in determining the solution of a differential equation of first order that's what makes me crazy

Answer 9

no he mentioned its a function of time see the question!!

Answer 10

y is a function of time

Answer 11

Then Y must be independent of t... !!

Answer 12

stop yelling @ me!! Ur hurting my feelings, bro :'(

Answer 13

awww.. come here you !!

Answer 14

-_-

Answer 15

wow i get a medol and abbot gets none.. yay yay :P

Answer 16

who the hell gave u a medal
i think your answer is wrong man here we're talking about partial derevative which is diffrent from derevating everything with respect to t

Answer 17

Is it the partial derivative of \[
F(u, t)
\]Or of \[
F(y(t),t)
\]?

Answer 18

the second one bro partial derevative with respect to t

Answer 19

Well it was ambiguous, and \(F(u, t) \neq F(y(t),t)\)

Answer 20

and so ?

Answer 21

if both functions are time dependent.. then really there is no difference between partial and normal derivatives.. HELL why would you even DO a partial derivative doesn't even make any sense :-/

Answer 22

-_- it makes sense when u'r solving a diferential equation of first order first you find F(y,t) then you derevative partially with respect to y or t and force it to be equal to the equation you have left so yeh u need a partial derevative

Answer 23

give me a differential equation!!

Answer 24

Like @Mashy is saying, \(F(y(t),t)\) doesn't have a partial derivative because it is not a multiple variable function. It's the OUTPUT of \(F(u, t)\) which happens to be a single variable function.

Answer 25

well, that escalated quickly.

Answer 26

lol :D

Answer 27

M(y,t)dt+N(y,t)dy=0 where M=df/dt and N=df/dy

Answer 28

Whoa, now we're doing ODE's?

Answer 29

thats called as EXACT FORM right?

Answer 30

Correct, Mashy :)

Answer 31

wio i didnt catch up with what you said i mean here we got 2 variable well not two variables y is a function of time and hell

Answer 32

well then in that.. case.. i have forgetting how to do it :D

Answer 33

This is escalating too damn quickly.

Answer 34

it's like i have holes in mathematics so when i advance i fall into some of them

Answer 35

holes in mathematics.. lol funny :D

Answer 36

Can you give us \(M, N\)?

Answer 37

he already mentioned.. df/dy and df/dt!

Answer 38

Well, you want to find the PD for either M or N first. I think it's ∂M/∂x and ∂N/∂y ?

Answer 39

it doesn't matter what M and N are the only restriction is that M and N are the partial derevatives of a certain function F(y,t)

Answer 40

wow i did this 4 years back!! ..

Answer 41

so i don't remember much :-/

Answer 42

sadly

Answer 43

Well, you have to check that they are exact first for this to work.

Answer 44

Otherwise, you are wasting time trying to do furhter work than is necessary.

Answer 45

\[
\begin{array}{rcl}
M(y,t)dt+N(y,t)dy &=& 0 \\
M(y,t)dt &=& -N(y,t)dy
\end{array}
\]
I would try integrating \(M\) with respect to \(t\) and then differentiating with respect to \(y\).

Answer 46

wait
can i post a picture ?

Answer 47

yup i can wait

Answer 48

really?!? of yourself?? we are in the middle of math here!

Answer 49

here check tht

Answer 50

i'm too handsome :p

Answer 51

so i have heard :P

Answer 52

i mean .. i did.. just now :P.. from you!!

Answer 53

so any idea about what i posted?

Answer 54

What it looks like you're trying to do in that step is find a integrating factor to make the solution exact?

Answer 55

yup

Answer 56

wait wait wait we found the integrating factor we're in the middle of finding F

Answer 57

see where's the arrow pointing

Answer 58

I'm not quite sure what you did there, I'm used to doing it a different method..

Answer 59

tht's the teacher's doing not me :3

Answer 60

oh well i'm off now i'll check again after half an hour or so thanks ppl

Answer 61

i'm back people any news,