What is the solution for y^{''}-9y^'+18y=e^{e^{-3x}} ?

The solution for y^{''}-9y^'+18y=e^{e^{-3x}} is y=c_{1}e^{6x}+c_{2}e^{3x}+(e^{6x+e^{-3x}})/9

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y^{''}-9y^'+18y=e^{e^{-3x}}

y^{^{''}} - 9 y^{^{'}} + 18 y = e^{e^{- 3 x}}

y = c_{1} e^{6 x} + c_{2} e^{3 x} + \frac{e ^{6 x + e^{- 3 x}}}{9}

Solution steps

y^{''} (x) - 9 y^{'} (x) + 18 y = e^{e^{- 3 x}}

Solve linear ODE:

y = c_{1} e^{6 x} + c_{2} e^{3 x} + \frac{e ^{6 x + e^{- 3 x}}}{9}

y = c_{1} e^{6 x} + c_{2} e^{3 x} + \frac{e ^{6 x + e^{- 3 x}}}{9}

\int - sin (- x + 2) d x

\int 4 y^{3} x d y

\int \frac{a ^{2} - x ^{2}}{2} d x

f (x) = \frac{x + 1810}{70 ( x + 2 )}

What is the solution for y^{''}-9y^'+18y=e^{e^{-3x}} ?
The solution for y^{''}-9y^'+18y=e^{e^{-3x}} is y=c_{1}e^{6x}+c_{2}e^{3x}+(e^{6x+e^{-3x}})/9