What is the solution for y^{''}+3y^'-10y=3xe^{-2x} ?

The solution for y^{''}+3y^'-10y=3xe^{-2x} is y=c_{1}e^{2x}+c_{2}e^{-5x}-(e^{-2x}x)/4+1/48 e^{-2x}

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y^{''}+3y^'-10y=3xe^{-2x}

y^{^{''}} + 3 y^{^{'}} - 10 y = 3 x e^{- 2 x}

y = c_{1} e^{2 x} + c_{2} e^{- 5 x} - \frac{e ^{- 2 x} x}{4} + \frac{1}{48} e^{- 2 x}

Solution steps

y^{''} (x) + 3 y^{'} (x) - 10 y = 3 x e^{- 2 x}

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- 5 x} - \frac{e ^{- 2 x} x}{4} + \frac{1}{48} e^{- 2 x}

y = c_{1} e^{2 x} + c_{2} e^{- 5 x} - \frac{e ^{- 2 x} x}{4} + \frac{1}{48} e^{- 2 x}

y^{^{''}} - 2 y^{^{''}} - 3 y = 9 x + 9

y^{^{'''}} - 9 y^{^{''}} + 27 y^{^{'}} - 27 y = 7 e^{3 x}

y^{^{''}} + 16 y = 50 e^{3} t, y (0) = 0, y^{^{'}} (0) = 0

4 y^{^{''}} - 8 y^{^{'}} + 5 y = e^{t} tan (\frac{t}{2})

(D^{2} + 1) y = 3 sin (2 t) - 2 cos (2 t)

What is the solution for y^{''}+3y^'-10y=3xe^{-2x} ?
The solution for y^{''}+3y^'-10y=3xe^{-2x} is y=c_{1}e^{2x}+c_{2}e^{-5x}-(e^{-2x}x)/4+1/48 e^{-2x}