What is the solution for xyy^'=1-x^2 ?

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

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xyy^'=1-x^2

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

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

Solution steps

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

Solve separable ODE:

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

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

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