What is the solution for y^{''}-6y^'+8y=4e^{e-2x} ?

The solution for y^{''}-6y^'+8y=4e^{e-2x} is y=c_{1}e^{4x}+c_{2}e^{2x}+(e^e)/6 e^{-2x}

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y^{''}-6y^'+8y=4e^{e-2x}

y^{^{''}} - 6 y^{^{'}} + 8 y = 4 e^{e - 2 x}

y = c_{1} e^{4 x} + c_{2} e^{2 x} + \frac{e ^{e}}{6} e^{- 2 x}

Solution steps

y^{''} (x) - 6 y^{'} (x) + 8 y = 4 e^{e - 2 x}

Solve linear ODE:

y = c_{1} e^{4 x} + c_{2} e^{2 x} + \frac{e ^{e}}{6} e^{- 2 x}

y = c_{1} e^{4 x} + c_{2} e^{2 x} + \frac{e ^{e}}{6} e^{- 2 x}

y^{^{''}} - 4 y^{^{'}} + 4 y = - 80 sin (3 x) - 23 cos (3 x)

\frac{d ^{2}}{d t ^{2}} (θ (t)) + 4 \cdot θ (t) = csc^{2} (2 t)

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

y^{^{'''}} + y^{^{''}} = 12 x^{5}

y^{^{''}} - 6 y^{^{'}} + 9 y = x^{- 3} e^{3 x}

What is the solution for y^{''}-6y^'+8y=4e^{e-2x} ?
The solution for y^{''}-6y^'+8y=4e^{e-2x} is y=c_{1}e^{4x}+c_{2}e^{2x}+(e^e)/6 e^{-2x}