What is the solution for y^'=(1+y^2)^{0.5} ?

The solution for y^'=(1+y^2)^{0.5} is y=-1/(2e^{t+c_{1)}}+(e^{t+c_{1}})/2

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y^'=(1+y^2)^{0.5}

y^{^{'}} = (1 + y^{2})^{0.5}

y = - \frac{1}{2 e ^{t + c_{1}}} + \frac{e ^{t + c_{1}}}{2}

Solution steps

y^{'} (t) = (1 + y^{2})^{0.5}

Solve separable ODE:

y = - \frac{1}{2 e ^{t + c_{1}}} + \frac{e ^{t + c_{1}}}{2}

y = - \frac{1}{2 e ^{t + c_{1}}} + \frac{e ^{t + c_{1}}}{2}

\frac{d y}{d x} (ln ∣ x + 8 ∣)^{6} = y

x \frac{d y}{d x} = 2 x^{3} y + y

\frac{d y}{d t} = \frac{3 y}{20} - \frac{y ^{2}}{1900}

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

\frac{d y}{1 + y} = \frac{d x}{1 + x}

What is the solution for y^'=(1+y^2)^{0.5} ?
The solution for y^'=(1+y^2)^{0.5} is y=-1/(2e^{t+c_{1)}}+(e^{t+c_{1}})/2