Question

how do i solve the equation [x^2 + (2x^2)y - 4y^3 = xy]. using y' = dy/dx? urgent i got a test on this in 4 hours

Answer 1

by solve what do you mean? there are lots of pairs of numbers (x,y) satisfying that, do you mean differentiate?

Answer 2

do you want to fing y' ?

Answer 3

find

Answer 4

what do you want us to do

Answer 5

the question was to find y' = dy/dx of the equation by implicit diffrentation

Answer 6

differentiate on both sides
use product rule where ever necessary
the answer would be
\[2x+4xy+2x ^{2}y \prime-12y ^{2}y \prime=xy \prime+y\]
now solve this for \[y \prime=(y-2x-4xy)/(2x ^{2}-12y ^{2}-x)\]

Answer 7

then find y' using implicit differentiation ...
http://www.youtube.com/watch?v=sL6MC-lKOrw

Answer 8

?

Answer 9

explain the final parts please..

Answer 10

??

Answer 11

dont get it

Answer 12

in the video or in the above problem??

Answer 13

the problem

Answer 14

fist step .. differentiate
second step ... take y' common <--- separate y' on one side!!

Answer 15

(y* dy/dx - 2 - 4y*dy/dx) / (4x-24*dy/dx-1) ??

Answer 16

??????

Answer 17

first off we will differentiate:

\[x^2 + 2x^2y - 4y^3 = xy\]
the first term:
\[\frac{d}{dx} (x^2) = 2x\]
the second term uses the product rule:
\[\frac{d}{dx} ( 2x^2 y) = 4xy + 2x^2 \frac{dy}{dx}\]

the third term uses the chain rule:
\[\frac{d}{dx} ( -4y^3) = -12y^2 \frac{dy}{dx}\]

on the other side we have another product rule:

\[\frac{d}{dx} ( xy )= y + x \frac{dy}{dx}\]

so differentiating the equation
\[x^2 + 2x^2y - 4y^3 = xy\]

we get
\[2x + 4xy +2x^2 \frac{dy}{dx} - 12x^2 \frac{dy}{dx} = y + x \frac{dy}{dx}\]

Answer 18

are you ok with that?

Answer 19

i'll continue, if anything is confusing just ask and i'll explain more :)

Answer 20

we now want to isolate dy/dx

so just as if we were isolating anything else, we want to collect "like" terms, factorising and stuff:

\[2x + 4xy +2x^2 \frac{dy}{dx} - 12x^2 \frac{dy}{dx} = y + x \frac{dy}{dx}\]

\[2x^2 \frac{dy}{dx} - 12x^2 \frac{dy}{dx} - x \frac{dy}{dx} = y -2x - 4xy\]\[\frac{dy}{dx} (2x^2 - 12x^2 - x ) =( y -2x - 4xy)\]\[\frac{dy}{dx} =\frac{ y -2x - 4xy}{2x^2 - 12x^2 - x }\]

Answer 21

warning: mistake!

Answer 22

that -12x^2 should be a -12y^2

Answer 23

\[\frac{dy}{dx} =\frac{ y -2x - 4xy}{2x^2 - 12y^2 - x }\]

Answer 24

hate typing