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Question

Find the particular solution of the differential equation x2dy=(2xy+y2) dx, given that y=1 when x=1.


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Solution

x2dy=(2xy+y2) dx

dydx=2xy+y2x2=2yx+(yx)2

which is homogeneous ....(i)

Put y=vx

v+xdvdx=2vx2+v2x2x2=2v+v2

xdvdx=v+v2

1v(v+1)dv=1xdx

1v1v+1dv=1xdx

log (v)log(v+1)=log x+log c

logvv+1=log cx

yy+x=cx ...(ii)

When y=1, x=1 then

11+1=c

c=12

Put the value of c in eq. (ii)

yy+x=12x

2y=xy+x2

x2+xy2y=0


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