What is the solution for y^{''}-8y^'+16y=2e^{4x} ?

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

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y^{''}-8y^'+16y=2e^{4x}

y^{^{''}} - 8 y^{^{'}} + 16 y = 2 e^{4 x}

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

Solution steps

y^{''} (x) - 8 y^{'} (x) + 16 y = 2 e^{4 x}

Solve linear ODE:

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

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

y^{^{''}} + 4 y^{^{'}} + 4 y = e^{x} cos (x)

y^{^{''}} + 4 y^{^{'}} + 13 y = 5 cos (2 t)

x^{^{''}} - 6 x^{^{'}} + 10 x = 30 t

y^{^{''}} - y^{^{'}} - 12 y = (x + 1) e^{x}

y^{^{''}} - 14 y^{^{'}} + 49 y = 35 x + 4

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