CBSE Class 12-science Answered

dy/dx = -y/(2(xy) - x )

Put y = vx

Therefore, dy/dx = v + xdv/dx

v + xdv/dx = -vx/(2(vx²) - x )

v + xdv/dx = -vx/(2xv - x )

v + xdv/dx = -v/(2v - 1 )

xdv/dx = [-v/(2v - 1 )] - v

xdv/dx = -2vv/(2v - 1)

Separating the variables,

(2v - 1)dv/(-2vv) = dx/x

Put 2v - 1 = z, and dv/v = dz

- 2zdz/(1+z)² = dx/x

Integrating both sides, by parts on LHS,

-2(-z/(1+z) + dz/(1+z)) = log x

2z/(1+z) -2log (1+z) = log x

2z/(1+z) = log (x(1+z)²)

Hence the solution is,

(2(y/x) - 1)/(y/x) = log 4y

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