What is the solution for 3y^{''}-5y^'+2y=6xe^{-2x} ?

The solution for 3y^{''}-5y^'+2y=6xe^{-2x} is y=c_{1}e^x+c_{2}e^{(2x)/3}+(e^{-2x}x)/4+17/96 e^{-2x}

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3y^{''}-5y^'+2y=6xe^{-2x}

3 y^{^{''}} - 5 y^{^{'}} + 2 y = 6 x e^{- 2 x}

y = c_{1} e^{x} + c_{2} e^{\frac{2 x}{3}} + \frac{e ^{- 2 x} x}{4} + \frac{17}{96} e^{- 2 x}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{x} + c_{2} e^{\frac{2 x}{3}} + \frac{e ^{- 2 x} x}{4} + \frac{17}{96} e^{- 2 x}

y = c_{1} e^{x} + c_{2} e^{\frac{2 x}{3}} + \frac{e ^{- 2 x} x}{4} + \frac{17}{96} e^{- 2 x}

y^{^{''}} + 2 y^{^{'}} + y = sin (x) + 7 cos (2 x)

y^{^{''}} + 4 y^{^{'}} + 4 y = (7 + x) e - 2 x

y^{^{''}} - y^{^{'}} - 6 y = e^{t}

\frac{d ^{2} y}{d x ^{2}} + 8 \frac{d y}{d x} + 25 y = 16 sin (x)

y^{^{''''}} - y = cos (x)

What is the solution for 3y^{''}-5y^'+2y=6xe^{-2x} ?
The solution for 3y^{''}-5y^'+2y=6xe^{-2x} is y=c_{1}e^x+c_{2}e^{(2x)/3}+(e^{-2x}x)/4+17/96 e^{-2x}