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

carson889 Group TitleBest ResponseYou've already chosen the best response.0
It is 2x + 8y  2.
 2 years ago

carson889 Group TitleBest 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)
 2 years ago

carson889 Group TitleBest 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, \[2(x+4y1)(1) = 2x + 8y  2\]
 2 years ago

Idealist Group TitleBest ResponseYou've already chosen the best response.0
2(x+4y1)=2x+8y2.
 2 years ago

gerryliyana Group TitleBest ResponseYou've already chosen the best response.0
ok thank all a lots i appreciate it :)
 2 years ago

Idealist Group TitleBest ResponseYou've already chosen the best response.0
No Problem.
 2 years ago

gerryliyana Group TitleBest 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???
 2 years ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.1
You need to use the chain rule, as this is implicit differentiation, not partial.
 2 years ago

gerryliyana Group TitleBest ResponseYou've already chosen the best response.0
what did u get it??
 2 years ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.1
So, are we trying to find \(y\) ?
 2 years ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.1
(As a function of \(x\) )
 2 years ago

gerryliyana Group TitleBest ResponseYou've already chosen the best response.0
it seems like http://openstudy.com/study#/updates/50a9b0e8e4b064039cbd16a2
 2 years ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.1
Yeah, so it's an actually differential equation which you have to solve. Here:
 2 years ago

LolWolf Group TitleBest 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.
 2 years ago

gerryliyana Group TitleBest ResponseYou've already chosen the best response.0
awesome..., thank u @LolWolf
 2 years ago

gerryliyana Group TitleBest ResponseYou've already chosen the best response.0
@LolWolf
 2 years ago

gerryliyana Group TitleBest ResponseYou've already chosen the best response.0
i tried to calculate on Maple 13, but, look at the picture below
 2 years ago

gerryliyana Group TitleBest ResponseYou've already chosen the best response.0
is not equal to (x+4y1)^2
 2 years ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.1
This is the equation I get, when solving the differential with Mathematica:
 2 years ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.1
In real form:
 2 years ago

gerryliyana Group TitleBest ResponseYou've already chosen the best response.0
it is right, isn't it?
 2 years ago
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