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

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

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

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

y = tan (\frac{1}{2} ln (x^{2} + 1) + c_{1})

Solution steps

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

Solve separable ODE:

y = tan (\frac{1}{2} ln (x^{2} + 1) + c_{1})

y = tan (\frac{1}{2} ln (x^{2} + 1) + c_{1})

x^{^{'}} + \frac{3}{200} x = 0.006

y^{2} \cdot d x - x \cdot y \cdot d y = x^{2} \cdot y \cdot d y

\frac{d p}{d t} = \frac{t \cdot cos ( t ^{2} )}{p ^{2}}

(y x^{2} - 4 y - x^{2} + 4) d x + d y = 0

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

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