What is the solution for y^'=(2xy)/(x^2-y^2),y(1)=1 ?

The solution for y^'=(2xy)/(x^2-y^2),y(1)=1 is y=1+sqrt(-x^2+1),y=1-sqrt(-x^2+1)

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y^'=(2xy)/(x^2-y^2),y(1)=1

y^{^{'}} = \frac{2 x y}{x ^{2} - y ^{2}}, y (1) = 1

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

Solution steps

y^{'} (x) = \frac{2 x y}{x ^{2} - y ^{2}}

Solve using substitution:

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

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

y = (x^{5} + 1)^{8}

\int 2 x + 2 x^{- 2} d x

\int \frac{x}{6 + π} d x

\frac{d}{d x} ((x^{2}) \cdot ln (x))

f (x) = \frac{x ^{2}}{9 + 2 x}

What is the solution for y^'=(2xy)/(x^2-y^2),y(1)=1 ?
The solution for y^'=(2xy)/(x^2-y^2),y(1)=1 is y=1+sqrt(-x^2+1),y=1-sqrt(-x^2+1)