Question

LGBADERIVATIVE:

\[\huge x^3 = (9x^2+y^2)C\]

Answer 1

you have to use implicit derivatives!

Answer 2

this should come out as \[3(3x^2 + y^2) dx - 2xydy = 0\]

Answer 3

that's my problem

Answer 4

that's supposed to be the answer?

Answer 5

yeah

Answer 6

C is just a constant?

Answer 7

yes. an arbitrary constant

Answer 8

OK. so we have
\[ \Large C=\frac{x^3}{9x^2+y^2} \]
then
\[ \Large 0=\frac{3x^2(9x^2+y^2)-x^3(18x+2y\frac{dy}{dx})}{?} \]
then
\[ \Large 0=27x^4+3x^2y^2-18x^4-2x^3y\frac{dy}{dx} \]

Answer 9

or could be it's\[x^3 = (9x^2 +y^2)C\]
that's wrong...

Answer 10

I get\[3x^2dx-18Cxdx-2Cydy=0\]\[3x^3dx-18Cx^2dx-2Cxydy=0\]\[3(9x^2+y^2)dx-18Cx^2dx-2Cxydy=0\]\[3((9-18C)x^2+y^2)dx-2Cxydy=0\]now what do we put for C ?

Answer 11

if you use the method outlined by @helder_edwin you will get the right answer.

Answer 12

how o.O

Answer 13

then
\[ \Large 0=9x^4+3x^2y^2-2x^3y\frac{dy}{dx} \]
then
\[ \Large 3x^2(3x^2+y^2)\,dx-2x^3y\,dy \]

Answer 14

@TuringTest i have no value for C...it was just a general solution haha

Answer 15

yeah, I couldn't simplify it, though that was my initial approach as well

Answer 16

lol wow cool @helder_edwin !!

Answer 17

well we have to get rid of it somehow, so a substitution seems to be the only way

Answer 18

seems helder figured it out already haha

Answer 19

um... is there an extra x^3 or am I misreading?

Answer 20

divide

Answer 21

\[\begin{align}
C&=\frac{x^3}{9x^2+y^2}\\
\therefore0&=\frac{3x^2}{9x^2+y^2}-\frac{x^3(18x+2yy')}{(9x^2+y^2)^2}
\end{align}\]which eventually simplifies to the answer...

Answer 22

oh ic

Answer 23

didn't notice the x^2

Answer 24

yup...seems turning thingies into differential equations is harder than i thought haha