What is the solution for y^'-4y^2e^{-4x}=0,y(0)=4 ?

The solution for y^'-4y^2e^{-4x}=0,y(0)=4 is y=-(4e^{4x})/(-4+3e^{4x)}

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y^'-4y^2e^{-4x}=0,y(0)=4

y^{^{'}} - 4 y^{2} e^{- 4 x} = 0, y (0) = 4

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

Solution steps

y^{'} (x) - 4 y^{2} e^{- 4 x} = 0

Solve separable ODE:

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

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

d x + y d y = x^{2} y d y

\frac{d y}{d x} = \frac{x}{y}, y = 2, x = 1

3 y^{2} \frac{d y}{d x} = (5 x)

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

\frac{d y}{d x} = \frac{( y + 3 ) ( x - 1 )}{( y - 2 ) ( x + 4 )}

What is the solution for y^'-4y^2e^{-4x}=0,y(0)=4 ?
The solution for y^'-4y^2e^{-4x}=0,y(0)=4 is y=-(4e^{4x})/(-4+3e^{4x)}