What is the solution for (dy)/(dx)=xe^{-2x} ?

The solution for (dy)/(dx)=xe^{-2x} is y= 1/4 (-2e^{-2x}x-e^{-2x})+c_{1}

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(dy)/(dx)=xe^{-2x}

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

y = \frac{1}{4} (- 2 e^{- 2 x} x - e^{- 2 x}) + c_{1}

Solution steps

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

Substitute

\frac{d y}{d x}

with

y^{'} (x)

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

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

y = \int x e^{- 2 x} d x

\int x e^{- 2 x} d x = \frac{1}{4} (- 2 e^{- 2 x} x - e^{- 2 x}) + c_{1}

y = \frac{1}{4} (- 2 e^{- 2 x} x - e^{- 2 x}) + c_{1}

\int \frac{2}{x + 3} d x

\frac{d}{d x} (\frac{x ^{2} + 3}{( x - 1 ) ^{3} + ( x + 1 ) ^{3}})

(6, - 5), (2, 3)

f (x) = (\frac{x - 3}{x + 3})^{2}, at x = - 2

What is the solution for (dy)/(dx)=xe^{-2x} ?
The solution for (dy)/(dx)=xe^{-2x} is y= 1/4 (-2e^{-2x}x-e^{-2x})+c_{1}