Question

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

Answer 1

solve \(\frac{ dy }{ dx } = (x + 4y -1)^{2}\)

Answer 2

It is 2x + 8y - 2.

Answer 3

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)

Answer 4

Or using chain rule: \[2(x+4y-1)^{1}(\frac{ d }{ dx }(x+4y-1))\]
\[\frac{ d }{ dx } = 1\]
Therfore,
\[2(x+4y-1)(1) = 2x + 8y - 2\]

Answer 5

2(x+4y-1)=2x+8y-2.

Answer 6

ok thank all a lots i appreciate it :)

Answer 7

No Problem.

Answer 8

@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???

Answer 9

You need to use the chain rule, as this is implicit differentiation, not partial.

Answer 10

what did u get it??

Answer 11

So, are we trying to find \(y\) ?

Answer 12

(As a function of \(x\) )

Answer 13

it seems like http://openstudy.com/study#/updates/50a9b0e8e4b064039cbd16a2

Answer 14

Yeah, so it's an actually differential equation which you have to solve.
Here:

Answer 15

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.

Answer 16

awesome..., thank u @LolWolf

Answer 17

Sure thing

Answer 18

@LolWolf

Answer 19

i tried to calculate on Maple 13, but, look at the picture below

Answer 20

is not equal to (x+4y-1)^2

Answer 21

This is the equation I get, when solving the differential with Mathematica:

Answer 22

In real form:

Answer 23

ok...,

Answer 24

it is right, isn't it?