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

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

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

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

y = c_{1} e^{2 (2 + 3) x} + c_{2} e^{2 (2 - 3) x} - e^{2 x}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 (2 + 3) x} + c_{2} e^{2 (2 - 3) x} - e^{2 x}

y = c_{1} e^{2 (2 + 3) x} + c_{2} e^{2 (2 - 3) x} - e^{2 x}

y^{^{''}} - 6 y^{^{'}} + 9 y = 16 e^{3 x}

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

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

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

y^{^{''''}} + 9 y^{^{''}} = sec^{2} (3 t)

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