Question

Verify 2xydx+(x^2-y)dy=0 -2x^2y+y^2=1

Answer 1

welcome 2 os

Answer 2

thank you, can you please help me with that equation? i really have no clue of what i have to do

Answer 3

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Answer 4

this is a example.

http://www.enotes.com/homework-help/how-solvethis-problem-3y-2-2xy-dx-2xy-x-2-dy-0-308165

Answer 5

go 2 this site and it will help

Answer 6

thanks, ill read it and if i have any question can i ask you?

Answer 7

so should i write it as -2xy/(x^2-y)=dy/dx?

Answer 8

yes

Answer 9

and then how do i get the substitution the example is telling to do next?

Answer 10

If you are not using differential equations, then you can solve the problem with implicit derivation.

Take \[F(x,y)=-2x^2y+y^2-1=0\]Then
\[dF=\frac{\partial F}{\partial x}dx+\frac{\partial F}{\partial y}dy=0\]
\[\frac{dy}{dx}=-\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial y}}\]

And the result follow from the last step.

Answer 11

So the partial derivative of F(x,y) in x is -4x and in y is -2x^2+2y=0?
Then i have \[-(-4x)/(-2x^2+2y)\]?

Answer 12

Im sorry to be such a mess! :(

Answer 13

You should have,

\[\frac{\partial F}{\partial x}=-4xy\]\[\frac{\partial F}{\partial y}=2y-2x^2\]Then
\[\frac{dy}{dx}=\frac{4xy}{2y-2x^2}=\frac{2xy}{y-x^2}\] And
\[2xydx+(x^2-y)dy=0\]

Answer 14

Thank you so much, sorry to not understand things so fast

Answer 15

Don't worry, all of us have a long way to learn. ;)

Answer 16

Can i ask you an extra question of another doubt i have? its not related to this problem

Answer 17

Of course.

Answer 18

Okay, so i have \[me ^{mx}+2e ^{mx}=0\], how do i clear the m?

Answer 19

well i know its with ln, but the ln goes outside the \[me ^{mx}\] or inside?

Answer 20

Well, I would do the following,
\[(m+2)e^{mx}=0\Rightarrow m=-2\]

Answer 21

oh boy, how could i not see it? haha now i really made a fool of myself.

Answer 22

Don't worry, remember the exponential could not reach the 0 value, it is an asymptote.

Answer 23

Okay, thank you so much for your help

Answer 24

You're welcome. One thing more, the link from previous answers (@kathy0514) is not correct, because the equation in your problem is not homogeneus. However the solved equation from the example is homogeneus.

Answer 25

oh okay, well i didnt get the part that it tells me to find the substitution, so i kind of got stuck in there.

Answer 26

Still, you can solve it by using differential equations. The equation is exact, so you can solve using the method from this video,
http://www.youtube.com/watch?v=bwASJWS8ltM
Follow the link only if you have taken the lessons of exact differential equations.

Answer 27

not yet, well see that topic next week, so i guess ill wait a little bit more

Answer 28

Ok, then problem solved ;)

Answer 29

perfect, thank you for your time! and have a great day!

Answer 30

Same to you, ;)