Question

answered question

Answer 1

\[\frac{ dM }{ dy } =\frac{ dN }{ dx }\]

Answer 2

hartnn can u help me

Answer 3

yeah,
y' = dy/dx

so, dy/dx =3y^2x^2
now bring it in the required form, by separating dy and dx

Answer 4

M dx+ Ndy = 0

Answer 5

how did you know that harnn

Answer 6

know what ?
y' = dy/dx ?

Answer 7

yes .. u right

Answer 8

y' is just another notation for dy/dx

Answer 9

ok harntt I see

Answer 10

http://www.purplemath.com/learning/viewtopic.php?f=14&t=124

Answer 11

but how can i prove that they r exactly deffirential equation

Answer 12

you mean *exact* differential equation ?

Answer 13

yes

Answer 14

i don't think it is exact DE ,are u sure question is correct ?

Answer 15

yes

Answer 16

im so stressed in this question >.<

Answer 17

ok, let me try

dy/dx = 3y^2x^2
dy - (3y^2x^2) dx = 0 this form will lead us to say that equation is not exact.

but let me try
(y^-2) dy - 2x^2 dx = 0 (this will lead to DE being exact)

comparing it with Mdx+Ndy= 0
here, M = -2x^2
N = y^-2

so, dM/dy = ... ?
dN/dx =... ?

Answer 18

can you find dM/dy when M = -2x^2 ?

Answer 19

0 ?

Answer 20

its actually, \(\large \dfrac{\partial M }{\partial y}\)

Answer 21

yes! 0 is correct! :)
what about dN/dx ???

Answer 22

is that right ?

Answer 23

yup

Answer 24

when?

Answer 25

0 again

Answer 26

now,
can you find dN/dy when N = y^-2 ?
yes 0 is correct again!

Answer 27

so, dM/dy = dN/dx
hence the DE is exact :)

Answer 28

any doubts ?

Answer 29

(y^-2) dy - 2x^2 dx = 0 (this will lead to DE being exact) whre did u get this?

Answer 30

by dividing the entire equation by y^2

Answer 31

you seperated the x an y?

Answer 32

dy/dx = 3y^2x^2

dy - 3y^2 x^2 dx = 0
dividing by y^2

dy/y^2 - 3y^2/y^2 x^2dx = 0

(y^-2) dy -3x^2 dx =0
got this ?

Answer 33

yes, i tried to separate them like that :)

Answer 34

thanks :D

Answer 35

welcome ^_^