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

The solution for (dy)/(dx)y=x^x is y=sqrt(integral of 2x^x),y=-sqrt(integral of 2x^x)

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(dy)/(dx)y=x^x

\frac{d y}{d x} y = x^{x}

y = \int 2 x^{x} d x, y = - \int 2 x^{x} d x

Solution steps

\frac{d y}{d x} y = x^{x}

Solve Bernoulli ODE:

y = \int 2 x^{x} d x, y = - \int 2 x^{x} d x

y = \int 2 x^{x} d x, y = - \int 2 x^{x} d x

8 x^{3} - 18 x^{2} - 6

\int - 2 x + 18 x^{2} - 4 x^{3} + 16 d x

x \to 3 lim (4 x^{2} - 2 x - 6)

y = \frac{x}{x + 9}, (1, 0.1)

x \to \infty lim (- 5 x^{2})

What is the solution for (dy)/(dx)y=x^x ?
The solution for (dy)/(dx)y=x^x is y=sqrt(integral of 2x^x),y=-sqrt(integral of 2x^x)