What is the solution for y^{''}-8y^'+16y=12e^{4x} ?

The solution for y^{''}-8y^'+16y=12e^{4x} is y=c_{1}e^{4x}+c_{2}xe^{4x}+6x^2e^{4x}

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y^{''}-8y^'+16y=12e^{4x}

y^{^{''}} - 8 y^{^{'}} + 16 y = 12 e^{4 x}

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

Solution steps

y^{''} (x) - 8 y^{'} (x) + 16 y = 12 e^{4 x}

Solve linear ODE:

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

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

y^{^{''}} + 2 y^{^{'}} - 8 y = 3 e^{- 3 x} - e^{- x}

x^{^{''}} + 1.1875 x^{^{'}} + 0.375 = 0

y^{^{''}} - y^{^{'}} - 2 y = 4, x^{2} y (0) = 1, y^{^{'}} (0) = 4

y^{^{''}} + 4 y^{^{'}} + 5 y = 4 + 12 t - 5 t^{2}

y^{^{''}} + 7 y^{^{'}} = 196 sin (7 t) + 588 cos (7 t)

What is the solution for y^{''}-8y^'+16y=12e^{4x} ?
The solution for y^{''}-8y^'+16y=12e^{4x} is y=c_{1}e^{4x}+c_{2}xe^{4x}+6x^2e^{4x}