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anonymous
 3 years ago
answred
anonymous
 3 years ago
answred

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0bernoulli equation or homogeneous

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0can you help me @.Sam.

.Sam.
 3 years ago
Best ResponseYou've already chosen the best response.0\[xdx + ( y 2x)dy= 0\] \[xdx=(y2x)dy \\ \frac{x}{y2x}=\frac{dy}{dx}\]  Try using y=vx Then \[\frac{dy}{dx}=v+x\frac{dv}{dx}\]

.Sam.
 3 years ago
Best ResponseYou've already chosen the best response.0Replace the y and \(\frac{dy}{dx}\) to v and the one given above \[\frac{x}{vx2x}=v+x\frac{dv}{dx} \\ \frac{1}{2v}=v+x\frac{dv}{dx}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the numerator can break it for the denomentor?

.Sam.
 3 years ago
Best ResponseYou've already chosen the best response.0Then set all v into the other side \[\frac{12v+v^2}{(2v)x}=\frac{dv}{dx} \\ \frac{(2v)x}{12v+v^2}=\frac{dx}{dv} \\ \int\limits \frac{(2v)}{12v+v^2} dx=\int\limits \frac{1}{x}dx\]

.Sam.
 3 years ago
Best ResponseYou've already chosen the best response.0Should be dv \[\int\limits\limits \frac{(2v)}{12v+v^2} dv=\int\limits\limits \frac{1}{x}dx\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i think you get stuck here though...no way to solve for "v" the integral gives a fraction and a log

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i think we can use here synthetic division

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@melmel , no that only works if the "2v" was in denominator

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0http://www.wolframalpha.com/input/?i=integral+%282v%29%2F%28v%5E22v%2B1%29

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0turns out you can't solve for "y" directly http://www.wolframalpha.com/input/?i=y%27%28x%29+%3D+x%2F%282xy%28x%29%29

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0can you help me to solve and to prove that?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@.sam, already did but you cant go any further just sub in y/x for v

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so what is my final answer?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i dunno depends on the instructions...ther is no "y =" answer

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0CAN YO HELP ME @zepdrix

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0coefficients by division?
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