What is the solution for e^{2y}y^'=1-e^{2y} ?

The solution for e^{2y}y^'=1-e^{2y} is y=(ln(-e^{-2t-c_{1}}+1))/2

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e^{2y}y^'=1-e^{2y}

e^{2 y} y^{^{'}} = 1 - e^{2 y}

y = \frac{ln ( - e ^{- 2 t - c_{1}} + 1 )}{2}

Solution steps

e^{2 y} y^{'} (t) = 1 - e^{2 y}

Solve separable ODE:

y = \frac{ln ( - e ^{- 2 t - c_{1}} + 1 )}{2}

y = \frac{ln ( - e ^{- 2 t - c_{1}} + 1 )}{2}

y = \frac{3}{x}, y = 4 - x

n = 0 \sum \infty (\frac{2}{9})^{n}

(1, 4), (- 9, - 10)

\int p^{6} ln (p) d p

\frac{1}{x ^{9}}

What is the solution for e^{2y}y^'=1-e^{2y} ?
The solution for e^{2y}y^'=1-e^{2y} is y=(ln(-e^{-2t-c_{1}}+1))/2