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

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

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

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

x = c_{1} e^{- 4 t} + c_{2} t e^{- 4 t} - \frac{1}{8} cos (4 t)

Solution steps

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

Solve linear ODE:

x = c_{1} e^{- 4 t} + c_{2} t e^{- 4 t} - \frac{1}{8} cos (4 t)

x = c_{1} e^{- 4 t} + c_{2} t e^{- 4 t} - \frac{1}{8} cos (4 t)

y^{^{''}} - 6 y^{^{'}} + 9 y = t, y (0) = 0, y (0) = 1

y^{^{''}} - 2 y^{^{'}} - 3 y = 3 x - 4

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

y^{^{''}} + 16 y = 34 e^{- 2 x} cos (4 x)

y^{^{'''}} - y^{^{''}} - y^{^{'}} + y = 9 e^{- t} + 2

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