What is the solution for y^'+3y=xe^{-2x} ?

The solution for y^'+3y=xe^{-2x} is y= x/(e^{2x)}-1/(e^{2x)}+(c_{1})/(e^{3x)}

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y^'+3y=xe^{-2x}

y^{^{'}} + 3 y = x e^{- 2 x}

y = \frac{x}{e ^{2 x}} - \frac{1}{e ^{2 x}} + \frac{c _{1}}{e ^{3 x}}

Solution steps

y^{'} (x) + 3 y = x e^{- 2 x}

Solve linear ODE:

y = \frac{x}{e ^{2 x}} - \frac{1}{e ^{2 x}} + \frac{c _{1}}{e ^{3 x}}

y = \frac{x}{e ^{2 x}} - \frac{1}{e ^{2 x}} + \frac{c _{1}}{e ^{3 x}}

What is the solution for y^'+3y=xe^{-2x} ?
The solution for y^'+3y=xe^{-2x} is y= x/(e^{2x)}-1/(e^{2x)}+(c_{1})/(e^{3x)}