What is the solution for (dy)/(dx)=(xy)/(ln(y)),y>1 ?

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

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(dy)/(dx)=(xy)/(ln(y)),y>1

\frac{d y}{d x} = \frac{x y}{ln ( y )}, y > 1

y = e^{x^{2} + c_{1}}, y = e^{- x^{2} + c_{1}}

Solution steps

\frac{d y}{d x} = \frac{x y}{ln ( y )}

Solve separable ODE:

y = e^{x^{2} + c_{1}}, y = e^{- x^{2} + c_{1}}

y = e^{x^{2} + c_{1}}, y = e^{- x^{2} + c_{1}}

(2 y^{2} + 3 y) d x + 2 x y d y = 0

y^{^{'}} = \frac{cos ( x )}{2 y}, y (0) = - 1

(x^{3} + 5 x^{3} y^{4}) d x + e^{x^{4}} y^{3} d y = 0

\frac{d y}{d x} = - \frac{6}{9} x^{5} y^{9}

y^{^{'}} = e^{x^{2}} 64 - y^{2}

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