What is the solution for y^'-2 y/(coth(2x))=(-2)/(coth(2x)) ?

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

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y^'-2 y/(coth(2x))=(-2)/(coth(2x))

y^{^{'}} - 2 \frac{y}{coth ( 2 x )} = \frac{- 2}{coth ( 2 x )}

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

Solution steps

y^{'} (x) - 2 \frac{y}{coth ( 2 x )} = \frac{- 2}{coth ( 2 x )}

Solve separable ODE:

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

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

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

e^{y} d x - x^{2} d y = 0

\frac{d y}{d x} = \frac{2 x}{- 8 y - 8}, y (0) = - 8

f^{^{'}} (x) = 9 f (x)

\frac{d y}{d x} y = (2 x + 5) (x^{3} - x^{8}) (x^{- 7})

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