solve differential equation (y^2 -2yx^2) dx + (2xy^2- x^2) dy=0
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Let v=
x
y
∴y=vx and dy=xdv+vdx
∴(x
2
+y
2
)dx−2xydy=0 becomes
(x
2
+v
2
x
2
)dx−2vx
2
(xdv+vdx)=0
∴(1+v
2
)dx=2v(xdv+vdx)
∴(1−v
2
)dx=2vxdv
∴
x
dx
=
1−v
2
2vdv
Integrating both sides, we have
logx=−log(1−v
2
)+c where c is an arbitrary constant
∴logx=log(
1−v
2
k
) where k is a constant greater than zero
∴k=x(1−v
2
)
∴k=x(1−
x
2
y
2
)
i.e. x
2
−y
2
=kx
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