Question

Hi everyone! Since an equation is exact if dM/dy = dN/dy, then how do I show if -f(x)dx+g(y)dy=0 is exact? Not sure how to show this in general terms. Thanks! :o)

Answer 1

I meant dM/dy = dN/dx

Answer 2

is f(x) a fnction of x alone ?

Answer 3

if so, the derivative with respect to `y` would be 0, yes ?

Answer 4

uhm...well...the deriv of x with respect to y...yes it would be zero I suppose

Answer 5

similarly the derivative of g(y) with respect to `x` would be 0

since they both are equal to 0, they are equal to each other.

Answer 6

interesting!...I didn't think that f(x) meant that "x" is actually the function...thinking of it this way, then yes they would both be zero! Thanks Ganeishi! :o)

Answer 7

I have such a hard time spelling your name! X0/

Answer 8

I'm not so sure if I understood your question correctly hmm

Answer 9

can you provide the details of full problem??

Answer 10

Well the whole question is simply prove that -f(x)dx+g(y)dy=0 is an exact equation

Answer 11

Ohkay.. then f(x) is a function of "x" alone
no y terms in f(x)

Answer 12

M = -f(x)
N = g(y)

dM/dy = d/dy (-f(x)) = 0
dN/dx = d/dy (g(y)) = 0


that means dM/dy = dN/dx = 0
so the given equation is exact.

Answer 13

got it! can you reason me through one more?

Answer 14

wil try, ask..

Answer 15

i need to confirm whether or not y=tan(x) is a solution to dy/dx=1+y^2 on interval -pi/2,pi/2

Answer 16

when I solve for y, i get y=arctan(x+c)

Answer 17

is that correct?

Answer 18

supposedly y=tan(x) is a solution on that interval but I can't seem to prove it

Answer 19

:o(

Answer 20

I think the confirmation is you finding y' given y=tan(x)
and seeing if the equation dy/dx=1+y^2 holds on the given interval

Answer 21

if y=tan(x) then y'=?

Answer 22

uhm...nuts....isn't that sec^2(x) ?

Answer 23

\[y'=1+y^2\\ \sec^2(x)=1+\tan^2(x) \text{ is this equation true ? } \]

Answer 24

lemme think a moment

Answer 25

hint: think Py...rean

Answer 26

yes it's true

Answer 27

I purposely let out some letters there

Answer 28

Ok also about the equation you found it should have been arctan(y)=x+c
where in this case they have c as 0
arctan(y)=x
so then y=tan(x) which is what we wanted to show was a solution

Answer 29

But when I think confirmation I think about the first way I talked about
just showing it is a solution instead of finding the solution

Answer 30

wait...one second...lemme say something...one sec

Answer 31

darn it...I just drew out the problem and it disappeared...one sec...sorry :o(

Answer 32

openstudy has been doing this thing to me where when I type things the page goes up while I'm typing so I can't see what I'm typing
and that is annoying
I don't know if that is why you accidentally deleted it or not
but some times I wish openstudy was a person so it can hear me yell at it

Answer 33

sorry...my pc kicked me off...lemme try again

Answer 34

okay, it's working now...one sec...thanks for being patient with me! :o)

Answer 35

ok waiting