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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{1}{6} y^{^{''}} + 6 y = 3 tan (6 t) - \frac{1}{3} e^{4 t}

y^{^{''}} - 4 y^{^{'}} + 8 y = x e^{x}

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

y^{^{''}} - 5 y^{^{'}} + 4 y = 8 e^{x}, y (0) = 1, y^{^{'}} (0) = 1

y^{^{''}} - 2 y^{^{'}} + 2 y = x + 1, y (0) = 3, y^{^{'}} (0) = 0

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