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

The solution for (1+x^2)y^{''}=2x(y^')^2 is y=integral of-1/(ln(x^2+1)+c_{1)}

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

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

y = \int - \frac{1}{ln ( x ^{2} + 1 ) + c _{1}} d x

Solution steps

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

Solve using substitution:

y = \int - \frac{1}{ln ( x ^{2} + 1 ) + c _{1}} d x

y = \int - \frac{1}{ln ( x ^{2} + 1 ) + c _{1}} d x

y^{^{'}} + \frac{1.93}{293 + t} \cdot y = 2.7249

\frac{\partial}{\partial x} (\frac{7 x - 5 y}{7 x + 5 y})

\frac{( s + 1 )}{6 s ^{2} + 7 s + 2}

x \to 2 lim (1 + 2 x e^{3 x})

x \to 0 - lim (∣ sin (x) ∣^{x})

What is the solution for (1+x^2)y^{''}=2x(y^')^2 ?
The solution for (1+x^2)y^{''}=2x(y^')^2 is y=integral of-1/(ln(x^2+1)+c_{1)}