What is the solution for y^'=(x-y)/(x+y),y(0)=1 ?

The solution for y^'=(x-y)/(x+y),y(0)=1 is y=-x+sqrt(2x^2+1)

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y^'=(x-y)/(x+y),y(0)=1

y^{^{'}} = \frac{x - y}{x + y}, y (0) = 1

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

Solution steps

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

Solve exact ODE:

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

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

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

\frac{d}{d x} (e^{x^{4} + x})

x^{2} + 3 x + 5

\frac{\partial}{\partial y} (x + 6 y)

\int x^{2} + 3 e^{x} + C d x

What is the solution for y^'=(x-y)/(x+y),y(0)=1 ?
The solution for y^'=(x-y)/(x+y),y(0)=1 is y=-x+sqrt(2x^2+1)