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Solve dy/dx = (x + 4y -1)^2

Mathematics
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solve \(\frac{ dy }{ dx } = (x + 4y -1)^{2}\)
It is 2x + 8y - 2.
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+4y-1)

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Other answers:

Or using chain rule: \[2(x+4y-1)^{1}(\frac{ d }{ dx }(x+4y-1))\] \[\frac{ d }{ dx } = 1\] Therfore, \[2(x+4y-1)(1) = 2x + 8y - 2\]
2(x+4y-1)=2x+8y-2.
ok thank all a lots i appreciate it :)
No Problem.
@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???
You need to use the chain rule, as this is implicit differentiation, not partial.
what did u get it??
So, are we trying to find \(y\) ?
(As a function of \(x\) )
Yeah, so it's an actually differential equation which you have to solve. Here:
Define (all in terms of x): \[ v=4y+x\implies v'=4y'+1 \]So: \[ \frac{v'-1}{4}=(v-1)^2\implies\\ \frac{v'}{4v^2-8v+5}=1 \]Integrating both sides: \[ \int \frac{v'}{4v^2-8v+5}\;dx=\int 1\;dx=\\ \frac{\tan^{-1}(2v-2)}{2}=x+C \]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.
awesome..., thank u @LolWolf
Sure thing
i tried to calculate on Maple 13, but, look at the picture below
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is not equal to (x+4y-1)^2
This is the equation I get, when solving the differential with Mathematica:
ok...,
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it is right, isn't it?

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