Question

just checking if im on the right track...i was asked to get solution for \(3xydx + (x^2 + y^2)dy = 0\) i used y = vx and came up with \[\frac{dx}{x} + \frac{(1+v^2)dx}{4c + v^3}\]

Answer 1

that's dv on the right side not dx

Answer 2

\[\frac{dy}{dx}=\frac{-3xy}{x^2+y^2}\]divide by x^2\[\frac{dy}{dx}=\frac{-3 \frac{y}{x}}{1+(\frac{y}{x})^2}\]
let y/x=v
y=vx
y'=v+v'x
\[v+\frac{dv}{dx}x=\frac{-3v}{1+v^2}\]\[\frac{dv}{dx}x=\frac{-3v-v(1+v^2)}{1+v^2}\]\[\frac{1+v^2}{-3v-v(1+v^2)}dv=\frac{dx}{x}\]

Answer 3

just add =0 on ur answer ull get it right

Answer 4

Josh:)

Answer 5

yay! thanks lilyna

Answer 6

ur welcome ewgeee:D

Answer 7

one more thing...how do you integrate \[\int \frac{(1+v^2)dv}{4v + v^3}\] the integral of the denom is 3v^2 + 4v and im sure that it's near

Answer 8

or do you use partail fractions??

Answer 9

yweah u can use partial fraction

Answer 10

stay with me for a while...ill try to get it :S

Answer 11

A = 1/16
B = -1/4
C = 17/16

those right?

Answer 12

hey

Answer 13

am i right? @pyrrolysinesynthesis ?

Answer 14

@eliassaab is my pf correct?

Answer 15

A/ v + ( Bv + C ) / ( v² + 4)
= A ( v² + 4 ) + ( Bv + C ) v
( A + B ) v² + Cv + 4A = v² + 1
-> C = 0, A = 1/4, B = 3/4