Question

Solve ordinary differential equation
xydy=(y^2+x)dx Solve ordinary differential equation
xydy=(y^2+x)dx @Mathematics

Answer 1

D[xy dy = (y^2+x)dx]

y = 2y y'

well, its not an exact is it

Answer 2

u don't have to write it all. I got udu=(1/x^2)dx (u=y/x). what now?

Answer 3

i spose the ^2 makes this nonlinear?

Answer 4

they aint homogenous either are they

Answer 5

im still starting these so I aint no expert yet ;)

Answer 6

you done bernoullis yet?

Answer 7

yes, but this is not bernoullis

Answer 8

yeah, i didnt think so either; but it would have been nice if it was lol

xy dy = (y^2+x) dx

xy y' = y^2+x

y' = (y^2+x)/xy

y' = y/x +1/y

hmmmm

Answer 9

y' - (1/x)y = y^(-1)

you sure we cant do a bernie on it?

Answer 10

well, I suppose that there's not just one way to solve this :)

Answer 11

http://www.wolframalpha.com/input/?i=y%27+%3D+y%2Fx+%2B+1%2Fy

i got some idea, but its prolly not correct

Answer 12

hahha, I used wolphram too :D

Answer 13

*wolfram

Answer 14

yeah, a the moment im at a loss

Answer 15

[y' - (1/x)y = y^(-1)]*y

y y' - (1/x)y^2 = 1

v = y^2, v'=2y y', y y' = v'/2

(1/2) v' - (1/x)v = 1

v' - (2/x)v = 2

e^(-2ln|x|) [v' - (2/x)v = 2]

e^(-2ln|x|)v = 2 int [e^(-2ln|x|)] dx

e^(-2ln|x|)v = 2 int [x^-2] dx

e^(-2ln|x|)v = -2/x + C

v = -2x^2/x + C x^2

v = -2x + C x^2; maybe

but thats just me dabbling .....

Answer 16

v = y^2

y^2 = -2x + C x^2

y = sqrt(-2x + Cx^2)

y = sqrt(x(-2 + Cx))

y = sqrt(x) sqrt( Cx -2)

which matches the wolfs