What is the solution for x^{''}+8x^'+16x=2sin(4t) ?

The solution for x^{''}+8x^'+16x=2sin(4t) is x=c_{1}e^{-4t}+c_{2}te^{-4t}-1/16 cos(4t)

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x^{''}+8x^'+16x=2sin(4t)

x^{^{''}} + 8 x^{^{'}} + 16 x = 2 sin (4 t)

x = c_{1} e^{- 4 t} + c_{2} t e^{- 4 t} - \frac{1}{16} cos (4 t)

Solution steps

x^{''} (t) + 8 x^{'} (t) + 16 x = 2 sin (4 t)

Solve linear ODE:

x = c_{1} e^{- 4 t} + c_{2} t e^{- 4 t} - \frac{1}{16} cos (4 t)

x = c_{1} e^{- 4 t} + c_{2} t e^{- 4 t} - \frac{1}{16} cos (4 t)

y^{^{''}} - 3 y^{^{'}} + 2 y = 7 e^{3 x}

y^{^{''}} - 2 y^{^{'}} - 3 y = 4 x - 5 + 6 x e^{3 x}

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

y^{^{''}} - 2 y^{^{'}} + 9 y = 2 cos (x)

y^{^{''}} + 4 y^{^{'}} + 4 y = (5 + x) e^{- 2 x}

What is the solution for x^{''}+8x^'+16x=2sin(4t) ?
The solution for x^{''}+8x^'+16x=2sin(4t) is x=c_{1}e^{-4t}+c_{2}te^{-4t}-1/16 cos(4t)