What is the solution for yy^'=x(y^2+1) ?

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

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yy^'=x(y^2+1)

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

y = e^{x^{2} + c_{1}} - 1, y = - e^{x^{2} + c_{1}} - 1

Solution steps

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

Solve separable ODE:

y = e^{x^{2} + c_{1}} - 1, y = - e^{x^{2} + c_{1}} - 1

y = e^{x^{2} + c_{1}} - 1, y = - e^{x^{2} + c_{1}} - 1

\frac{d}{d x} (\frac{x e ^{x} - e ^{x}}{x ^{2}})

x \to 0 lim (3 x + 1)

\frac{4}{x - 7}

\frac{\partial}{\partial x} (x^{2} + 2 x y - y^{2} - 8 x)

f (t) = - 4 t^{2} + 16 t + 9

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