Question

HELP ME IN DIFFERENTIA EQUATION TOPIC IS EXACT

Answer 1

PROBLEM IS\[(\frac{ x }{ x^2+y^2 }) dx +( \frac{ y }{ x^2+y^2 }) dy = 0\]

Answer 2

Take partial derivative in terms of x and take partial derivative in terms of y and if it's the same then it is an exact equation and we can go further.

Answer 3

so i need to multiply both side by x^2+y^2

Answer 4

no no no... you have to take partial derivative
example

\[x^2+y^2 \] the partial derivative in terms of x is 2x

Answer 5

well I think you could try it...idk lack of sleep and going through all sorts of stuff I'm burned out

Answer 6

i think we can multiply by x^2+y^2

Answer 7

Seems like you lose a bunch of information doing that :o
Hmm I dunno.

But you should just do your partials :D
\[\large\rm \frac{\partial }{\partial y}\left(\frac{x}{x^2+y^2}\right)=?\]

Answer 8

so we will have xdx + ydy = 0

Answer 9

i think i got it now \[[(\frac{ x }{ x^2+y^2 })dx + (\frac{ y }{ x^2+y^2 }dy)] = 0\] (x^2+y^2)
\[\int\limits xdx + \int\limits ydy = 0\]
\[x^2+y^2= c\]

Answer 10

but the problem is idont know if it is exact

Answer 11

in getting if it is exact do i need to separate the \[\frac{ x }{ x^2 } and \frac{ 1 }{ y }\]

Answer 12

sec I'm trying to brush up on my exact equations c: lol
don't really remember all of this stuff

Answer 13

So if you have a function in two variables, \(\large\rm f(x,y)\),
then it's derivative is given by,\[\large\rm f'(x,y)=f_x\frac{dx}{dx}+f_y\frac{dy}{dx}\]Which we can write as,\[\large\rm f'(x,y)=f_x+f_y\frac{dy}{dx}\]And if it's exact, then,\[\large\rm f'(x,y)=f_x+f_y\frac{dy}{dx}=0\]Which we choose to write this way,\[\large\rm f_x dx+f_y dy=0\]Somethingggg like that I think.
Maybe it's coming back to me.

Answer 14

We check for exactness first,\[\large\rm \frac{\partial}{\partial y}\left(\frac{x}{x^2+y^2}\right)=\frac{-2xy}{(x^2+y^2)^2}\]Do you understand how to find that partial?
Is that part confusing?

Answer 15

wait... we can get the derivative on the new equation that i got when i multiply by the denominatr \[\frac{ dM }{ dy } = d(x) and \frac{ dN }{ dx } = d(y)\]

Answer 16

derivative of x is 0 and derivative of y is 0 also so theyre exact, is that right?

Answer 17

You really shouldn't multiply both sides by (x^2+y^2).
You're losing so much information.
Let that go.

Answer 18

JUST USE QUOTIENT RULE AND TAKE PARTIAL DERIVATIVES FIRST TO TEST EXACTNESS!