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}+1/(e^2+3e+2)e^{ex}

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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} + \frac{1}{e ^{2} + 3 e + 2} e^{e 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} + \frac{1}{e ^{2} + 3 e + 2} e^{e x}

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

y^{^{''}} - 5 y^{^{'}} + 6 y = 2 e^{x}

(d^{2} - 1) \cdot y = x \cdot sin (x) + x^{2} \cdot e^{x}

so l v e f or d, d (x) = \frac{1}{36 ( x - 60 ) ^{2} - 100}

(d^{2} + 5 d + 4) y = sin^{3} (x)

(d + 4) y = x^{2} cos^{2} (x)

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}+1/(e^2+3e+2)e^{ex}