What is the solution for y^'+4y=xe^{-4x} ?

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

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y^'+4y=xe^{-4x}

y^{^{'}} + 4 y = x e^{- 4 x}

y = \frac{x ^{2}}{2 e ^{4 x}} + \frac{c _{1}}{e ^{4 x}}

Solution steps

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

Solve linear ODE:

y = \frac{x ^{2}}{2 e ^{4 x}} + \frac{c _{1}}{e ^{4 x}}

y = \frac{x ^{2}}{2 e ^{4 x}} + \frac{c _{1}}{e ^{4 x}}

\int_{- 1}^{2} x e^{x y} d y

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

\int 5 x^{7} d x

\frac{d}{d x} (x^{2} + z (x) (x)^{2})

\frac{d}{d x} (csc^{2} (3 x))

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