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

The solution for y^{''}+3y^'+2y=xe^x is y=c_{1}e^{-x}+c_{2}e^{-2x}+(e^xx)/6-5/36 e^x

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

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

y = c_{1} e^{- x} + c_{2} e^{- 2 x} + \frac{e ^{x} x}{6} - \frac{5}{36} e^{x}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} e^{- 2 x} + \frac{e ^{x} x}{6} - \frac{5}{36} e^{x}

y = c_{1} e^{- x} + c_{2} e^{- 2 x} + \frac{e ^{x} x}{6} - \frac{5}{36} e^{x}

y^{^{''}} - 4 y^{^{'}} - 21 y = 480 e^{- t}

y^{^{''}} - 9 y^{^{'}} + 20 y = x + 1

y^{^{''}} - 5 y^{^{'}} + 6 y = cos (2 x)

y^{^{''}} - 2 y^{^{'}} + y = e^{t}, y (0) = 0, y (0) = 5

y^{^{''}} + 16 y = 3 sec (4 x)

What is the solution for y^{''}+3y^'+2y=xe^x ?
The solution for y^{''}+3y^'+2y=xe^x is y=c_{1}e^{-x}+c_{2}e^{-2x}+(e^xx)/6-5/36 e^x