What is the solution for yy^'e^{y^2}=x-1 ?

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

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yy^'e^{y^2}=x-1

y y^{^{'}} e^{y^{2}} = x - 1

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

Solution steps

y y^{'} (x) e^{y^{2}} = x - 1

Solve separable ODE:

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

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

y^{^{'}} = - y cos (x)

y^{^{'}} - 2 t y^{3} = 0, y (2) = 1

y^{^{'}} = \frac{7 y}{x}

\frac{d y}{d x} = \frac{x 1 - y ^{2}}{y 2 x ^{2} + 1}

\frac{d y}{d x} = y (2 + sin (x))

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