What is the solution for xy+2ln(y)y^'=0 ?

The solution for xy+2ln(y)y^'=0 is y=e^{sqrt((-x^2+c_{1))/2}},y=e^{-sqrt((-x^2+c_{1))/2}}

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xy+2ln(y)y^'=0

x y + 2 ln (y) y^{^{'}} = 0

y = e^{\frac{- x ^{2} + c _{1}}{2}}, y = e^{- \frac{- x ^{2} + c _{1}}{2}}

Solution steps

x y + 2 ln (y) y^{'} (x) = 0

Solve separable ODE:

y = e^{\frac{- x ^{2} + c _{1}}{2}}, y = e^{- \frac{- x ^{2} + c _{1}}{2}}

y = e^{\frac{- x ^{2} + c _{1}}{2}}, y = e^{- \frac{- x ^{2} + c _{1}}{2}}

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e^{x} \cdot y \frac{d y}{d x} = e^{- y} + e^{- 2 x - y}

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What is the solution for xy+2ln(y)y^'=0 ?
The solution for xy+2ln(y)y^'=0 is y=e^{sqrt((-x^2+c_{1))/2}},y=e^{-sqrt((-x^2+c_{1))/2}}