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

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

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

y^{^{'}} = \frac{2 x}{y + 1}

y = - 1 + 2 (x^{2} + c_{1}) + 1, y = - 2 (x^{2} + c_{1}) + 1 - 1

Solution steps

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

Solve separable ODE:

y = - 1 + 2 (x^{2} + c_{1}) + 1, y = - 2 (x^{2} + c_{1}) + 1 - 1

y = - 1 + 2 (x^{2} + c_{1}) + 1, y = - 2 (x^{2} + c_{1}) + 1 - 1

\frac{d}{d x} (arcsech (x))

\frac{d}{d x} (\frac{1}{a + b \cdot ( \frac{x}{d} ) ^{c}})

\frac{d y}{d t} + 5 y = 80, y (1) = 100

\frac{d}{d x} (- x + 1)

\int \frac{1}{tan ( x ) sec ( x )} d x

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