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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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

\frac{x}{1 + x ^{2}}, at x = 0

\int 36 x e^{- 9 x} d x

\frac{2 x + 1}{x + 5} \cdot (3 x - 1)

e^{16 x}

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