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

The solution for y^'=(y^3)/(e^x),y(0)=-1 is y=-sqrt(-(e^x)/(-2+e^x))

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y^'=(y^3)/(e^x),y(0)=-1

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

y = - - \frac{e ^{x}}{- 2 + e ^{x}}

Solution steps

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

Solve separable ODE:

y = - - \frac{e ^{x}}{- 2 + e ^{x}}

y = - - \frac{e ^{x}}{- 2 + e ^{x}}

\frac{d y}{d x} = (4 - y)^{5}

(1 + x^{2}) \frac{d y}{d x} = 2 x y + x^{2} + x^{4}

\int \frac{1}{( 1 + e ^{x} )} d x

\frac{- 2}{( s + 1 ) ( s + 2 )}

x \to 0 lim (\frac{cos ( x ^{2} )}{x})

What is the solution for y^'=(y^3)/(e^x),y(0)=-1 ?
The solution for y^'=(y^3)/(e^x),y(0)=-1 is y=-sqrt(-(e^x)/(-2+e^x))