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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y, (d^{3} - d) y = 2 cos (t)

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

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

(d^{2} - 6 d + 9) y = 2 e^{3} x

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

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