What is the solution for (e^x-e^{-x})(dy)/(dx)=y^2 ?

The solution for (e^x-e^{-x})(dy)/(dx)=y^2 is y=-2/(ln(tanh(x/2))+c_{1)}

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(e^x-e^{-x})(dy)/(dx)=y^2

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

y = - \frac{2}{ln ( tanh ( \frac{x}{2} ) ) + c _{1}}

Solution steps

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

Solve separable ODE:

y = - \frac{2}{ln ( tanh ( \frac{x}{2} ) ) + c _{1}}

y = - \frac{2}{ln ( tanh ( \frac{x}{2} ) ) + c _{1}}

\frac{\partial}{\partial x} (- 10 x + 7 y^{5})

\frac{d}{d x} (\frac{1}{4 x ^{3}} - \frac{6}{5 x ^{4}})

\frac{d y}{d x} = (\frac{2 y + 5}{8 x + 7})^{2}

y^{^{'}} - 3 y = 3 e^{4 t}

2 x + y + (x + 2 y) \cdot y^{^{'}} = 0

What is the solution for (e^x-e^{-x})(dy)/(dx)=y^2 ?
The solution for (e^x-e^{-x})(dy)/(dx)=y^2 is y=-2/(ln(tanh(x/2))+c_{1)}