Question

need help expanding out the brackets please:
(4x^3 +4xy) times (1 - 2x^2 - 2y)^-1

Answer 1

if the 2nd polynomial is all to the -1 power it would look like this

\[(4x^3+4xy)*\frac{ 1 }{ (1-2x^2-2y) }\]

Answer 2

yep, but how do i expand it out?

Answer 3

\[\frac{ 4x^3+4xy }{ 1-2x^2-2y }\]

Answer 4

Im doing ODE's, it's currently:
y' = (4x^3 +4xy) /(1 - 2x^2 - 2y)
need it in the form:
y' + (P(x)) y = Q(x)

Answer 5

@Peter14 lil help if you got a sec?

Answer 6

see...

Answer 7

multiply the whole thing by (1−2x^2−2y)

Answer 8

y' = (4x^3 +4xy) /(1 - 2x^2 - 2y)

You may wish to observe this: \(y' = \dfrac{2x\cdot(2x^{2}+2y)}{1 - (2x^{2} + 2y)}\)

Answer 9

so y ' = 2x (2x^2 +2y)/1 - (2x (2x^2 +2y)/(2x^2 +2y)?

Answer 10

which equals :

y ' = 2x (2x^2 +2y) - 2x ...?

Answer 11

Have you considered substitution from your studies of integration?

Take a look at the derivative (dy/dx) of \(ln(1-(2x^{2} + 2y))\).

I'm not saying this leads directly to a solution. Just think on it.

Answer 12

no, sorry at the moment i'm only comfortable with solving from :
y' + (P(x)) y = Q(x)
and finding the integrating factor

Answer 13

thanks for the help tho