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

The solution for y^{''}+2y^'-8y=2e^{2x} is y=c_{1}e^{2x}+c_{2}e^{-4x}+(e^{2x}x)/3

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

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

y = c_{1} e^{2 x} + c_{2} e^{- 4 x} + \frac{e ^{2 x} x}{3}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- 4 x} + \frac{e ^{2 x} x}{3}

y = c_{1} e^{2 x} + c_{2} e^{- 4 x} + \frac{e ^{2 x} x}{3}

y^{^{''}} + y = cos (2 t), y (0) = 0, y^{^{'}} (0) = 1

3 y^{^{''}} + 5 y^{^{'}} + 2 y = t + e^{t}

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

y^{^{''}} - 8 y^{^{'}} + 16 y = 2 t e^{4 t}

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

What is the solution for y^{''}+2y^'-8y=2e^{2x} ?
The solution for y^{''}+2y^'-8y=2e^{2x} is y=c_{1}e^{2x}+c_{2}e^{-4x}+(e^{2x}x)/3