solve by homogeneous method dx/dy=x²+y²/2xy
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Answer:
Given, (x
2
+y
2
)dx−2xydy=0
⇒(x
2
+y
2
)dx=2xydy
⇒
dx
dy
=
2xy
x
2
+y
2
.... (i)
Let y=vx
Thus,
dx
dy
=v+x
dx
dv
Thus, v+x
dx
dv
=
2x(vx)
x
2
+(vx)
2
⇒v+x
dx
dv
=
2v
1+v
2
⇒x
dx
dv
=
2v
1+v
2
−v
⇒x
dx
dv
=
2v
1+v
2
−2v
2
⇒x
dx
dv
=
2v
1−v
2
⇒
x
dx
=
1−v
2
2v
dv
⇒
x
dx
−
1−v
2
2v
dv=0 .... (ii)
Integrating both sides, we have
logx+log(1−v
2
)=logC
⇒logx(1−v
2
)=logC
⇒x(1−v
2
)=C
⇒x(1−
x
2
y
2
)=C
⇒x(
x
2
x
2
−y
2
)=C
⇒x
2
−y
2
=Cx
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