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anonymous
 3 years ago
Solve dy/dx = (x + 4y 1)^2
anonymous
 3 years ago
Solve dy/dx = (x + 4y 1)^2

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

carson889
 3 years ago
Best ResponseYou've already chosen the best response.0Got 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)

carson889
 3 years ago
Best ResponseYou've already chosen the best response.0Or using chain rule: \[2(x+4y1)^{1}(\frac{ d }{ dx }(x+4y1))\] \[\frac{ d }{ dx } = 1\] Therfore, \[2(x+4y1)(1) = 2x + 8y  2\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok thank all a lots i appreciate it :)

anonymous
 3 years ago
Best 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???

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You need to use the chain rule, as this is implicit differentiation, not partial.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So, are we trying to find \(y\) ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0(As a function of \(x\) )

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it seems like http://openstudy.com/study#/updates/50a9b0e8e4b064039cbd16a2

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yeah, so it's an actually differential equation which you have to solve. Here:

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Define (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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0awesome..., thank u @LolWolf

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i tried to calculate on Maple 13, but, look at the picture below

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0This is the equation I get, when solving the differential with Mathematica:

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it is right, isn't it?
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