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

- anonymous

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- anonymous

option 2

- anonymous

since both are function of t!

- abb0t

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

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## More answers

- anonymous

yes

- abb0t

\[\frac{ ∂G }{ ∂t }\]

- anonymous

yes

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

y is a function of time

- anonymous

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

- abb0t

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

- anonymous

awww.. come here you !!

- anonymous

-_-

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

the second one bro partial derevative with respect to t

- anonymous

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

- anonymous

and so ?

- anonymous

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 :-/

- anonymous

-_- 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

- anonymous

give me a differential equation!!

- anonymous

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.

- abb0t

well, that escalated quickly.

- anonymous

lol :D

- anonymous

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

- abb0t

Whoa, now we're doing ODE's?

- anonymous

thats called as EXACT FORM right?

- abb0t

Correct, Mashy :)

- anonymous

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

- anonymous

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

- abb0t

This is escalating too damn quickly.

- anonymous

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

- anonymous

holes in mathematics.. lol funny :D

- anonymous

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

- anonymous

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

- abb0t

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

- anonymous

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)

- anonymous

wow i did this 4 years back!! ..

- anonymous

so i don't remember much :-/

- anonymous

sadly

- abb0t

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

- abb0t

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

- anonymous

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

- anonymous

wait
can i post a picture ?

- anonymous

yup i can wait

- anonymous

##### 1 Attachment

- anonymous

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

- anonymous

here check tht

- anonymous

i'm too handsome :p

- anonymous

so i have heard :P

- anonymous

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

- anonymous

so any idea about what i posted?

- abb0t

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

- anonymous

yup

- anonymous

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

- anonymous

see where's the arrow pointing

- abb0t

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

- anonymous

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

- anonymous

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

- anonymous

i'm back people any news,

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