Question

Solve (x-y)dx+xdy=0

Answer 1

y=ux dy=udu+xdx
dx=1 so dy=udu+xdx

I try to substitute. Is it ok if I leave out the dx?

Answer 2

y=ux dy=udu+xdx
dx=1 so dy=udu+x

Answer 3

You have to integrate no?

Answer 4

do I?

Answer 5

What topic of the course is this from?

Answer 6

homogeneous substitution

Answer 7

i got y=-xlnx+xC

Answer 8

(x-ux)dx+x(udu+x)=0
xdx-uxdx+xudu+x^2=0

work correct so far?

Answer 9

where the C above is a constant

Answer 10

tell me myininaya if my work is correct

Answer 11

i dont know the way you are doing it
im sorry

Answer 12

k. ty.

Answer 13

but i posted above how i got my answer if that helps

Answer 14

please visit site orlandoicc.org

Answer 15

why do you multiply by v?

Answer 16

i multiply by v (i choose this v such that we have vy'+v'y=(yv)')

Answer 17

if we can write the differential equation in this form y'+p(x)y=q(x) then we can multiply by v such that we have vy'+vpy=vq but we want this v so that v'=vp so we can write the next step as
(vy)'=vq

Answer 18

v'=pv
dv/dx=pv
1/v dv=p dx
lnv=p dx
so
we have v=e^(int p dx)

Answer 19

so we multiply whatever that v is on both sides of the equation then we can write the
(vy)'=vq
integrate both sides
vy=int(vq)+C
solve for y
y=1/v*int(vq) +C/v