anonymous
  • anonymous
Solve dy/dx = (x + 4y -1)^2
Mathematics
jamiebookeater
  • jamiebookeater
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
solve \(\frac{ dy }{ dx } = (x + 4y -1)^{2}\)
carson889
  • carson889
It is 2x + 8y - 2.
carson889
  • carson889
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)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

carson889
  • carson889
Or using chain rule: \[2(x+4y-1)^{1}(\frac{ d }{ dx }(x+4y-1))\] \[\frac{ d }{ dx } = 1\] Therfore, \[2(x+4y-1)(1) = 2x + 8y - 2\]
anonymous
  • anonymous
2(x+4y-1)=2x+8y-2.
anonymous
  • anonymous
ok thank all a lots i appreciate it :)
anonymous
  • anonymous
No Problem.
anonymous
  • anonymous
@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
  • anonymous
You need to use the chain rule, as this is implicit differentiation, not partial.
anonymous
  • anonymous
what did u get it??
anonymous
  • anonymous
So, are we trying to find \(y\) ?
anonymous
  • anonymous
(As a function of \(x\) )
anonymous
  • anonymous
Yeah, so it's an actually differential equation which you have to solve. Here:
anonymous
  • anonymous
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.
anonymous
  • anonymous
awesome..., thank u @LolWolf
anonymous
  • anonymous
Sure thing
anonymous
  • anonymous
anonymous
  • anonymous
i tried to calculate on Maple 13, but, look at the picture below
1 Attachment
anonymous
  • anonymous
is not equal to (x+4y-1)^2
anonymous
  • anonymous
This is the equation I get, when solving the differential with Mathematica:
anonymous
  • anonymous
anonymous
  • anonymous
ok...,
1 Attachment
anonymous
  • anonymous
it is right, isn't it?

Looking for something else?

Not the answer you are looking for? Search for more explanations.