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gerryliyanaBest ResponseYou've already chosen the best response.0
solve \(\frac{ dy }{ dx } = (x + 4y 1)^{2}\)
 one year ago

carson889Best ResponseYou've already chosen the best response.0
Got it by expanding the brackets: x^2 + 4yx  x + 4xy + 16y^2 4y x  4y + 1. Which simplifies to: x^2 + 8xy + 16y^2  2x 8y + 1. Taking the derivative of that with respect to x: 2x + 8y + 0  2  0 + 0 Which simplifies to 2x + 8y 2 or 2(x+4y1)
 one year ago

carson889Best ResponseYou've already chosen the best response.0
Or using chain rule: \[2(x+4y1)^{1}(\frac{ d }{ dx }(x+4y1))\] \[\frac{ d }{ dx } = 1\] Therfore, \[2(x+4y1)(1) = 2x + 8y  2\]
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.0
ok thank all a lots i appreciate it :)
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.0
@carson889: wait i mean \[\frac{ dy }{ dx } = (x+4y  1)^{}\] \((x+4y 1)^{2}\) is the result of \(\frac{ dy }{ dx }\) it isn't \[\frac{ d }{ dx} (x +4y 1)^{2}\] any idea???
 one year ago

LolWolfBest ResponseYou've already chosen the best response.1
You need to use the chain rule, as this is implicit differentiation, not partial.
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.0
what did u get it??
 one year ago

LolWolfBest ResponseYou've already chosen the best response.1
So, are we trying to find \(y\) ?
 one year ago

LolWolfBest ResponseYou've already chosen the best response.1
(As a function of \(x\) )
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.0
it seems like http://openstudy.com/study#/updates/50a9b0e8e4b064039cbd16a2
 one year ago

LolWolfBest ResponseYou've already chosen the best response.1
Yeah, so it's an actually differential equation which you have to solve. Here:
 one year ago

LolWolfBest ResponseYou've already chosen the best response.1
Define (all in terms of x): \[ v=4y+x\implies v'=4y'+1 \]So: \[ \frac{v'1}{4}=(v1)^2\implies\\ \frac{v'}{4v^28v+5}=1 \]Integrating both sides: \[ \int \frac{v'}{4v^28v+5}\;dx=\int 1\;dx=\\ \frac{\tan^{1}(2v2)}{2}=x+C \]Solving for \(v\): \[ v=\frac{\tan(2(x+C))+2}{2} \]Substituting \(y\): \[ y=\frac{\tan(2(x+C))2x+2}{8} \] That took forever.
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.0
awesome..., thank u @LolWolf
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.0
i tried to calculate on Maple 13, but, look at the picture below
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.0
is not equal to (x+4y1)^2
 one year ago

LolWolfBest ResponseYou've already chosen the best response.1
This is the equation I get, when solving the differential with Mathematica:
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.0
it is right, isn't it?
 one year ago
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