What is the solution for y^'=(e^{-x}-e^x)/(3+4y),y(0)=1 ?

The solution for y^'=(e^{-x}-e^x)/(3+4y),y(0)=1 is y=(-3+sqrt(-8e^x-8e^{-x)+65})/4

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y^'=(e^{-x}-e^x)/(3+4y),y(0)=1

y^{^{'}} = \frac{e ^{- x} - e ^{x}}{3 + 4 y}, y (0) = 1

y = \frac{- 3 + - 8 e ^{x} - 8 e ^{- x} + 65}{4}

Solution steps

y^{'} (x) = \frac{e ^{- x} - e ^{x}}{3 + 4 y}

Solve separable ODE:

y = \frac{- 3 + - 8 e ^{x} - 8 e ^{- x} + 65}{4}

y = \frac{- 3 + - 8 e ^{x} - 8 e ^{- x} + 65}{4}

y^{2} - 1 - (2 y + x y) y^{^{'}} = 0

\frac{d y}{d x} x^{3} 4 y = 5

y^{5} + 9 y^{^{'}} = 0

\frac{d y}{d x} = 4 e^{x - y}, y (0) = ln (6)

y^{^{'}} = - 2 x (y^{2} - 3 y + 2), y (0) = 3

What is the solution for y^'=(e^{-x}-e^x)/(3+4y),y(0)=1 ?
The solution for y^'=(e^{-x}-e^x)/(3+4y),y(0)=1 is y=(-3+sqrt(-8e^x-8e^{-x)+65})/4