Solve (x-y)dx+xdy=0

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Solve (x-y)dx+xdy=0

Mathematics
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y=ux dy=udu+xdx dx=1 so dy=udu+xdx I try to substitute. Is it ok if I leave out the dx?
y=ux dy=udu+xdx dx=1 so dy=udu+x
You have to integrate no?

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Other answers:

do I?
What topic of the course is this from?
homogeneous substitution
i got y=-xlnx+xC
(x-ux)dx+x(udu+x)=0 xdx-uxdx+xudu+x^2=0 work correct so far?
where the C above is a constant
tell me myininaya if my work is correct
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i dont know the way you are doing it im sorry
k. ty.
but i posted above how i got my answer if that helps
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why do you multiply by v?
i multiply by v (i choose this v such that we have vy'+v'y=(yv)')
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
v'=pv dv/dx=pv 1/v dv=p dx lnv=p dx so we have v=e^(int p dx)
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

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