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

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

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(dy)/(dx)=ln(x^y),y(1)=1

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

y = e^{- x + 1} x^{x}

Solution steps

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

Solve separable ODE:

y = e^{- x + 1} x^{x}

y = e^{- x + 1} x^{x}

(y^{2} + y) d x - x d y = 0, y (1) = 1

y^{^{'}} = - x^{5} y^{2}

\frac{d y}{d x} + \frac{y}{5} = y^{- 3}

\frac{d y}{d x} = - \frac{3 cos ( x )}{y}

(x + 2)^{2} \cdot \frac{d y}{d x} = 4 x y

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