What is the solution for (dx)/(dy)=(1+x^2)/(1-y^2) ?

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

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(dx)/(dy)=(1+x^2)/(1-y^2)

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

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

Solution steps

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

Solve separable ODE:

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

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

u^{^{'}} + n^{2} H (t) = - \frac{( - 1 ) ^{n} + 1}{π ( n ^{2} - 1 )}

\frac{d y}{d x} - 5 x = x y

y^{^{'}} - sin (x) \cdot y = 3 sin (x)

\frac{d y}{d t} = \frac{t ^{2}}{1 + y ^{2}}, y (0) = 1

y^{2} y^{^{'}} = (1 + y^{3}) cos (x)

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