What is the solution for y^'+8y=xe^{-8x} ?

The solution for y^'+8y=xe^{-8x} is y=(x^2)/(2e^{8x)}+(c_{1})/(e^{8x)}

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y^'+8y=xe^{-8x}

y^{^{'}} + 8 y = x e^{- 8 x}

y = \frac{x ^{2}}{2 e ^{8 x}} + \frac{c _{1}}{e ^{8 x}}

Solution steps

y^{'} (x) + 8 y = x e^{- 8 x}

Solve linear ODE:

y = \frac{x ^{2}}{2 e ^{8 x}} + \frac{c _{1}}{e ^{8 x}}

y = \frac{x ^{2}}{2 e ^{8 x}} + \frac{c _{1}}{e ^{8 x}}

64 - x^{2}

h \to 0 lim (- \frac{1}{x ( x + h )})

\frac{\partial}{\partial x} (e x^{3} - y^{5})

f (x) = x^{e^{x}}

\frac{d}{d x} (4^{2 x})

What is the solution for y^'+8y=xe^{-8x} ?
The solution for y^'+8y=xe^{-8x} is y=(x^2)/(2e^{8x)}+(c_{1})/(e^{8x)}