What is the solution for (dy)/(dx)=x*e^{-2x},y(0)=3 ?

The solution for (dy)/(dx)=x*e^{-2x},y(0)=3 is y= 1/4 (-2e^{-2x}x-e^{-2x})+13/4

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(dy)/(dx)=x*e^{-2x},y(0)=3

\frac{d y}{d x} = x \cdot e^{- 2 x}, y (0) = 3

y = \frac{1}{4} (- 2 e^{- 2 x} x - e^{- 2 x}) + \frac{13}{4}

Solution steps

\frac{d y}{d x} = x \cdot 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}

Apply initial conditions:

y = \frac{1}{4} (- 2 e^{- 2 x} x - e^{- 2 x}) + \frac{13}{4}

y = \frac{1}{4} (- 2 e^{- 2 x} x - e^{- 2 x}) + \frac{13}{4}

\int sin (n x) d x

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

x \to \frac{π}{2} + lim (tan (x))

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

What is the solution for (dy)/(dx)=x*e^{-2x},y(0)=3 ?
The solution for (dy)/(dx)=x*e^{-2x},y(0)=3 is y= 1/4 (-2e^{-2x}x-e^{-2x})+13/4