What is the solution for y^'=y^2e^{-t},y(0)= 1/2 ?

The solution for y^'=y^2e^{-t},y(0)= 1/2 is y=(e^t)/(1+e^t)

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y^'=y^2e^{-t},y(0)= 1/2

y^{^{'}} = y^{2} e^{- t}, y (0) = \frac{1}{2}

y = \frac{e ^{t}}{1 + e ^{t}}

Solution steps

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

Solve separable ODE:

y = \frac{e ^{t}}{1 + e ^{t}}

y = \frac{e ^{t}}{1 + e ^{t}}

\frac{d y}{d x} = \frac{e ^{8 y + x}}{x}, x > 0

y^{^{'}} = 8 x sec (y)

e^{x} tan (y) d x + (1 - e^{x}) sec^{2} (y) d y = 0

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

(5 x + 4) + (5 y - 5) y^{^{'}} = 0

What is the solution for y^'=y^2e^{-t},y(0)= 1/2 ?
The solution for y^'=y^2e^{-t},y(0)= 1/2 is y=(e^t)/(1+e^t)