anonymous
  • anonymous
Verify 2xydx+(x^2-y)dy=0 -2x^2y+y^2=1
Differential Equations
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
welcome 2 os
anonymous
  • anonymous
thank you, can you please help me with that equation? i really have no clue of what i have to do
goformit100
  • goformit100
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More answers

anonymous
  • anonymous
this is a example. http://www.enotes.com/homework-help/how-solvethis-problem-3y-2-2xy-dx-2xy-x-2-dy-0-308165
anonymous
  • anonymous
go 2 this site and it will help
anonymous
  • anonymous
thanks, ill read it and if i have any question can i ask you?
anonymous
  • anonymous
so should i write it as -2xy/(x^2-y)=dy/dx?
anonymous
  • anonymous
yes
anonymous
  • anonymous
and then how do i get the substitution the example is telling to do next?
John_ES
  • John_ES
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.
anonymous
  • anonymous
So the partial derivative of F(x,y) in x is -4x and in y is -2x^2+2y=0? Then i have \[-(-4x)/(-2x^2+2y)\]?
anonymous
  • anonymous
Im sorry to be such a mess! :(
John_ES
  • John_ES
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 \[2xydx+(x^2-y)dy=0\]
anonymous
  • anonymous
Thank you so much, sorry to not understand things so fast
John_ES
  • John_ES
Don't worry, all of us have a long way to learn. ;)
anonymous
  • anonymous
Can i ask you an extra question of another doubt i have? its not related to this problem
John_ES
  • John_ES
Of course.
anonymous
  • anonymous
Okay, so i have \[me ^{mx}+2e ^{mx}=0\], how do i clear the m?
anonymous
  • anonymous
well i know its with ln, but the ln goes outside the \[me ^{mx}\] or inside?
John_ES
  • John_ES
Well, I would do the following, \[(m+2)e^{mx}=0\Rightarrow m=-2\]
anonymous
  • anonymous
oh boy, how could i not see it? haha now i really made a fool of myself.
John_ES
  • John_ES
Don't worry, remember the exponential could not reach the 0 value, it is an asymptote.
anonymous
  • anonymous
Okay, thank you so much for your help
John_ES
  • John_ES
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.
anonymous
  • anonymous
oh okay, well i didnt get the part that it tells me to find the substitution, so i kind of got stuck in there.
John_ES
  • John_ES
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.
anonymous
  • anonymous
not yet, well see that topic next week, so i guess ill wait a little bit more
John_ES
  • John_ES
Ok, then problem solved ;)
anonymous
  • anonymous
perfect, thank you for your time! and have a great day!
John_ES
  • John_ES
Same to you, ;)

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