anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
option 2
anonymous
  • anonymous
since both are function of t!
abb0t
  • abb0t
\[F(y(t), t) = y \times G(t)\] ?

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anonymous
  • anonymous
yes
abb0t
  • abb0t
\[\frac{ ∂G }{ ∂t }\]
anonymous
  • anonymous
yes
anonymous
  • anonymous
if both G and Y are function of t.. then you have to use the product rule!
anonymous
  • 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
  • anonymous
no he mentioned its a function of time see the question!!
anonymous
  • anonymous
y is a function of time
anonymous
  • anonymous
Then Y must be independent of t... !!
abb0t
  • abb0t
stop yelling @ me!! Ur hurting my feelings, bro :'(
anonymous
  • anonymous
awww.. come here you !!
anonymous
  • anonymous
-_-
anonymous
  • anonymous
wow i get a medol and abbot gets none.. yay yay :P
anonymous
  • 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
  • anonymous
Is it the partial derivative of \[ F(u, t) \]Or of \[ F(y(t),t) \]?
anonymous
  • anonymous
the second one bro partial derevative with respect to t
anonymous
  • anonymous
Well it was ambiguous, and \(F(u, t) \neq F(y(t),t)\)
anonymous
  • anonymous
and so ?
anonymous
  • 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
  • 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
  • anonymous
give me a differential equation!!
anonymous
  • 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
  • abb0t
well, that escalated quickly.
anonymous
  • anonymous
lol :D
anonymous
  • anonymous
M(y,t)dt+N(y,t)dy=0 where M=df/dt and N=df/dy
abb0t
  • abb0t
Whoa, now we're doing ODE's?
anonymous
  • anonymous
thats called as EXACT FORM right?
abb0t
  • abb0t
Correct, Mashy :)
anonymous
  • 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
  • anonymous
well then in that.. case.. i have forgetting how to do it :D
abb0t
  • abb0t
This is escalating too damn quickly.
anonymous
  • anonymous
it's like i have holes in mathematics so when i advance i fall into some of them
anonymous
  • anonymous
holes in mathematics.. lol funny :D
anonymous
  • anonymous
Can you give us \(M, N\)?
anonymous
  • anonymous
he already mentioned.. df/dy and df/dt!
abb0t
  • abb0t
Well, you want to find the PD for either M or N first. I think it's ∂M/∂x and ∂N/∂y ?
anonymous
  • 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
  • anonymous
wow i did this 4 years back!! ..
anonymous
  • anonymous
so i don't remember much :-/
anonymous
  • anonymous
sadly
abb0t
  • abb0t
Well, you have to check that they are exact first for this to work.
abb0t
  • abb0t
Otherwise, you are wasting time trying to do furhter work than is necessary.
anonymous
  • 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
  • anonymous
wait can i post a picture ?
anonymous
  • anonymous
yup i can wait
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
really?!? of yourself?? we are in the middle of math here!
anonymous
  • anonymous
here check tht
anonymous
  • anonymous
i'm too handsome :p
anonymous
  • anonymous
so i have heard :P
anonymous
  • anonymous
i mean .. i did.. just now :P.. from you!!
anonymous
  • anonymous
so any idea about what i posted?
abb0t
  • abb0t
What it looks like you're trying to do in that step is find a integrating factor to make the solution exact?
anonymous
  • anonymous
yup
anonymous
  • anonymous
wait wait wait we found the integrating factor we're in the middle of finding F
anonymous
  • anonymous
see where's the arrow pointing
abb0t
  • abb0t
I'm not quite sure what you did there, I'm used to doing it a different method..
anonymous
  • anonymous
tht's the teacher's doing not me :3
anonymous
  • anonymous
oh well i'm off now i'll check again after half an hour or so thanks ppl
anonymous
  • anonymous
i'm back people any news,

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