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

The solution for y^'= 1/(1+y) is y=-1+sqrt(2(t+c_{1))+1},y=-sqrt(2(t+c_{1))+1}-1

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y^'= 1/(1+y)

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

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

Solution steps

y^{'} (t) = \frac{1}{1 + y}

Solve separable ODE:

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

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

x^{2} (1 - 3 x)

x \to 1 lim (- (3))

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

\int_{0}^{\infty} \frac{1}{2 ^{x}} d x

x \to 0 lim (x^{6} cos (\frac{5}{x}))

What is the solution for y^'= 1/(1+y) ?
The solution for y^'= 1/(1+y) is y=-1+sqrt(2(t+c_{1))+1},y=-sqrt(2(t+c_{1))+1}-1