Question

xy'=y+2*sqrt(xy)

Somewhere in there is a homogeneous equation yearning to be free, I can't find it though..

Answer 1

are you needing the integral?

Answer 2

The general solution, the ex bank i'm running though are all first order DQs.

normally, i'd v=y/x, but that root throws me off.

Answer 3

dont think of it as a root then. sqrt (x)= x^(1/2)

Answer 4

then dividing both sides, y'=y/x + 2(xy)^(1/2)/x

Answer 5

I'm attempting to format this a a homeogeneous FODQ but perhaps I don't know all the tricks of the trade

Answer 6

which can be rephrased.
y'=y*x^(-1)+2*y^(1/2)*x^(-1/2)

Answer 7

its easier looking at all powers rather than sqrts and fractions

Answer 8

so then that's 1/2 v ?
for the subsitution
\[v+v'x = v + 2 v^(1/2)\]

Answer 9

then I can solve by seperation of variables correct?

Answer 10

what are you setting as v?

Answer 11

I'm subbing v=y/x so y'=v +v'x

Answer 12

ok. yes that would be v^(1/2) then. you good from there?

Answer 13

appears so, thank you, now i'll just see if works out in the end

Answer 14

lol. good luck. simple things like looking at the same number in a different way does wonders.

Answer 15

cherrio,y=xln(x)^2 +xc

Answer 16

again - its hfodeq if y'=f(x,y) where f(x,xt) = f(1,t)