What is the solution for (e^x+e^{-x})y^'=y^2 ?

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

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(e^x+e^{-x})y^'=y^2

(e^{x} + e^{- x}) y^{^{'}} = y^{2}

y = - \frac{1}{arctan ( e ^{x} ) + c _{1}}

Solution steps

(e^{x} + e^{- x}) y^{'} (x) = y^{2}

Solve separable ODE:

y = - \frac{1}{arctan ( e ^{x} ) + c _{1}}

y = - \frac{1}{arctan ( e ^{x} ) + c _{1}}

\int x^{- 4} + 2 x + 3 d x

x \to 0 lim (\frac{2 1 ^{x} - 1}{x})

\int \frac{1}{2 x ( 1 + x )} d x

y^{^{'}} = \frac{2}{3} x 1 - 9 y^{2}

f (x) = x^{2} + 2, at x = 3

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