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Question

If the curve y=y(x) represented by the solution of the differential equation (2xy2y)dx+xdy=0, passes through the intersection of the lines, 2x3y=1 and 3x+2y=8, then |y(1)| is equal to

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Solution

Given,
(2xy2y)dx+xdy=0
dydx+2y2yx=0
1y2dydx+1y(1x)=2
Let, 1y=z
1y2dydx=dzdx
dzdx+z(1x)=2
I.F. =e1xdx=x
z(x)=2(x)dx=x2+c
xy=x2+c
As it passes through P(2, 1)
[Point of intersection of 2x3y=1 and 3x+2y=8]
21=4+c
c=2
xy=x22
Put x=1
1y=12=1
y(1)=1
|y(1)|=1

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