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}

\int \frac{1}{( x ^{2} - 64 ) ^{\frac{3}{2}}} d x

x \to 5 lim (c x^{2} + 6 x)

y = \frac{1}{2 x}

\int e^{\frac{x}{3} + 9} d x

x \to 2 lim (\frac{x + 7}{x ^{2} - 4})

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)}