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}

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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