What is the solution for y^{''}+3y^'+2y=12e^{-3x} ?

The solution for y^{''}+3y^'+2y=12e^{-3x} is y=c_{1}e^{-x}+c_{2}e^{-2x}+6e^{-3x}

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y^{''}+3y^'+2y=12e^{-3x}

y^{^{''}} + 3 y^{^{'}} + 2 y = 12 e^{- 3 x}

y = c_{1} e^{- x} + c_{2} e^{- 2 x} + 6 e^{- 3 x}

Solution steps

y^{''} (x) + 3 y^{'} (x) + 2 y = 12 e^{- 3 x}

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} e^{- 2 x} + 6 e^{- 3 x}

y = c_{1} e^{- x} + c_{2} e^{- 2 x} + 6 e^{- 3 x}

y^{^{''}} - 12 y^{^{'}} + 36 y = 30 x + 3

y^{^{''}} + y^{^{'}} - 2 y = u π (t) - u^{2} π (t), y (0) = 1

x^{^{''}} + 4 x^{^{'}} + 40 x = 20 sin (6 t)

y^{^{''}} - y^{^{'}} = 8

y^{^{''}} - y^{^{'}} = 3

What is the solution for y^{''}+3y^'+2y=12e^{-3x} ?
The solution for y^{''}+3y^'+2y=12e^{-3x} is y=c_{1}e^{-x}+c_{2}e^{-2x}+6e^{-3x}