What is the solution for yy^'=xe^{y^2} ?

The solution for yy^'=xe^{y^2} is y=sqrt(-ln(-x^2-c_{1))},y=-sqrt(-ln(-x^2-c_{1))}

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yy^'=xe^{y^2}

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

y = - ln (- x^{2} - c_{1}), y = - - ln (- x^{2} - c_{1})

Solution steps

y y^{'} (x) = x e^{y^{2}}

Solve separable ODE:

y = - ln (- x^{2} - c_{1}), y = - - ln (- x^{2} - c_{1})

y = - ln (- x^{2} - c_{1}), y = - - ln (- x^{2} - c_{1})

y^{^{'}} - x y = 0

\frac{d y}{d x} = \frac{2 y}{( x + 1 ) + ( x + 1 ) ^{\frac{5}{2}}}

y^{^{'}} + 4 y = 0, y (0) = cos (2 t), y (0) = sin (2 t)

\frac{d y}{d x} = \frac{cos ( x )}{2 y}

- y = k \frac{d y}{d x}

What is the solution for yy^'=xe^{y^2} ?
The solution for yy^'=xe^{y^2} is y=sqrt(-ln(-x^2-c_{1))},y=-sqrt(-ln(-x^2-c_{1))}