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xon1300
 2 years ago
Best ResponseYou've already chosen the best response.0use green's theorem ( http://en.wikipedia.org/wiki/Green%27s_theorem )

gerryliyana
 2 years ago
Best ResponseYou've already chosen the best response.0Seriously??? can you tell me how to solve it?

xon1300
 2 years ago
Best ResponseYou've already chosen the best response.0It's been a while since i took calculus 3

gerryliyana
 2 years ago
Best ResponseYou've already chosen the best response.0have you any idea, @CynosureEPR ?????

CynosureEPR
 2 years ago
Best ResponseYou've already chosen the best response.0I'm sorry =( I do know how to do it but I'm rushed with a programming assignment (worth 40% of grade) due at midnight =(

gerryliyana
 2 years ago
Best ResponseYou've already chosen the best response.0ok no problem @CynosureEPR :) good luck for u :)

gerryliyana
 2 years ago
Best ResponseYou've already chosen the best response.0@zepp have u any idea??

zepp
 2 years ago
Best ResponseYou've already chosen the best response.0Nope, I'm only at Calc II, sorry ):

gerryliyana
 2 years ago
Best ResponseYou've already chosen the best response.0ok np @zepp thank u for coming :)

mahmit2012
 2 years ago
Best ResponseYou've already chosen the best response.1dw:1354582292989:dw

ChmE
 2 years ago
Best ResponseYou've already chosen the best response.0\[(x2y +1) dx + (4x 3y 6) dy = 0\]\[(x2y +1) dx = (4x 3y 6) dy \]\[\frac{ dy }{ dx }=\frac{ (x2y+1) }{ (4x 3y 6) }\]

ChmE
 2 years ago
Best ResponseYou've already chosen the best response.0Do you know how to do separation of variables? http://en.wikipedia.org/wiki/Separation_of_variables

ChmE
 2 years ago
Best ResponseYou've already chosen the best response.0This looks like a diff eq question

ChmE
 2 years ago
Best ResponseYou've already chosen the best response.0nvm, I don't think we can use that here

gerryliyana
 2 years ago
Best ResponseYou've already chosen the best response.0then ???? how to solve y?? i really know separation variable, but in this case that method cannot work,

gerryliyana
 2 years ago
Best ResponseYou've already chosen the best response.0yes.., i think so.., we cannot use that in this case.., :)

ChmE
 2 years ago
Best ResponseYou've already chosen the best response.0My teacher skipped partial derivatives and that looks like what is needed for this problem. I'm going to tapout. Good luck

gerryliyana
 2 years ago
Best ResponseYou've already chosen the best response.0i see.., wonderful site for me.., thank u @ChmE,

sirm3d
 2 years ago
Best ResponseYou've already chosen the best response.0solve the system of equation \[\large \left(\begin{matrix}x2y+1=0 \\ 4x3y6=0\end{matrix}\right)\]

sirm3d
 2 years ago
Best ResponseYou've already chosen the best response.0when you get the solution, say x0 and y0, use u = x  x0 and v = y  y0. try this @gerryliyana

sirm3d
 2 years ago
Best ResponseYou've already chosen the best response.0did you get this DE? \[\large (u2v)du + (4u3v)dv=0\] x changed to u, y changed to v, the constants +1 and 6 disappeared.

gerryliyana
 2 years ago
Best ResponseYou've already chosen the best response.0\[\left( \frac{ x  2y + 1 }{ 4x 3y  6 } \right) \times \left[ \frac{ 4 }{ 1 } \right]\] \[\left( \frac{ 4x  8y + 4 }{ 4x  3y 6 } \right) = 0\] \[\left( \frac{ 4x  8y = 4 }{ 4x 3y= 6 } \right)\]  __ \[8y (3y) = 10 \rightarrow y = 2\] \[4x  3(2) = 6 \rightarrow x = 3\]

sirm3d
 2 years ago
Best ResponseYou've already chosen the best response.0on the right track. now, u = x  3 and v = y  2

gerryliyana
 2 years ago
Best ResponseYou've already chosen the best response.0OMG..., my connection is lost, and I lost all my type, :(

sirm3d
 2 years ago
Best ResponseYou've already chosen the best response.0hahaha. enter them a few equations at a time.

sirm3d
 2 years ago
Best ResponseYou've already chosen the best response.0don't worry. you are not the first one.
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